All language subtitles for Calculus-1-Full-College-Course_en

This video is about working with rational\n

usually with variables in it, something like\n

rational expression. In this video, we'll\n

and dividing rational expressions, and simplifying\n

to lowest terms. Recall, if you have a fraction\n

over 45, we can reduce it to lowest terms\n

denominator and then canceling common factors.\n

our fraction reduces to seven over 15. If\n

variables and add to lowest terms, we proceed\n

that's three times x plus two, and then factor\n

plus two times x plus two, we could also write\n

the common factors. And we're left with three\n

of writing that rational expression. Now next,\n

that if we multiply two fractions with just\n

and multiply the denominators. So in this\n

times five or 8/15. If we want to divide two\n

we can rewrite it as multiplying by the reciprocal\n

we get four fifths times three halves, and\n

could reduce that fraction to six fifths,\n

product or quotient of two rational expressions\n

to divide to rational expressions. So instead,\n

this flipping and multiplying. And now we\n

the denominators. It might be tempting at\n

the numerator and the denominator. But actually,\n

and factored even more completely. That way,\n

to cancel the common factors. So let's factor\n

x times x plus one, and x squared minus 16.\n

x plus four times x minus four, the denominator\n

it over. And now we can cancel common factors\n

x minus four. This is our final answer. Adding\n

complicated because we first have to find\n

is an expression that both denominators divided\n

use the least common denominator, which is\n

In this example, if we just want a common\n

is 90 because both six and 15 divided evenly\n

the best way to do that is to factor the two\n

times five, and then put together only the\n

our numbers, so if we just use two times three\n

times three will divide it and three times\n

to get a denominator any smaller because we\n

order to ensure both these numbers divided,\n

we can rewrite each of our fractions in terms\n

to get a 30 in the denominator, so I'm going\n

by the factors that are missing from the current\n

denominator of 30. For the second fraction,\n

multiply by two over two, I can rewrite this\n

a common denominator, I can just subtract\n

I can reduce this to three squared over two\n

for finding the sum of two rational expressions\n

process. First, we have to find the least\n

the two denominators. So 2x plus two factors\n

that's a difference of two squares. So that's\n

least common denominator, I'm going to take\n

that each of these divides into, so I need\n

and I need the factor x minus one, I don't\n

need to have at one time. And so I will get\n

one times x minus one, I'm not going to bother\n

to leave it in factored form to help me simplify\n

expressions by multiplying by whatever's missing\n

common denominator. So what I mean is, I can\n

the 2x plus two is two times x plus one, I'll\n

that compared to the least common denominator,\n

I multiply the numerator and the denominator\n

but I can't get away with just multiplying\n

I have to multiply by it on the numerator\n

by one and a fancy form and not changing the\n

the second rational expression. I'll write\n

easier to see what's missing from the denominator.\n

to my least common denominator is just the\n

the denominator by two. Now I can rewrite\n

becomes three times x minus one over two times\n

five times two over two times x plus 1x minus\n

So I can just add together my numerators.\n

two times x plus 1x minus one. I'd like to\n

is to leave the denominator in factored form.\n

so that I can add things together. So I get\n

1x minus one, or 3x plus seven over two times\n

factor. And there's therefore no factors that\n

As much as it can be. This is my final answer.

In this video, we saw how to simplify rational\n

canceling common factors. We also saw how\n

the numerator and multiplying the denominator\n

and multiplying, and how to add and subtract\n

of the least common denominator. This video\n

rate of change. These are topics that are\n

For function y equals f of x, like the function\n

lie. that stretches between two points on\n

this x value has a, and this x value as B.\n

have an x value of a, and a y value given\n

B and Y value, f of b. Now the average rate\n

a to b can be defined as the slope have the\n

A and B, F of B. In symbols, that slope m\n

y over the change in x, which is the difference\n

the difference in x coordinates b minus a.\n

put this in context, if for example, f of\n

time in years, then f of b minus f of a represents\n

tree grows. And B minus A represents a difference\n

rate of change is the amount the tree grows\n

it grows 10 inches in two years, that would\n

per year would be its average rate of change\n

the average rate of change for the function\n

the average rate of change is f of four minus\n

is the square root of four of one's the square\n

one over three or 1/3. Instead of calling\n

this time, I'm going to call the first location,\n

The idea is that h represents the horizontal\n

x axis. In this notation, if I want to label\n

it'll have an x coordinate of x and a y coordinate\n

coordinate of x plus h and a y coordinate\n

simply the average rate of change using this\n

represents the average rate of change of a\n

x plus h. Equivalent way, the difference quotient\n

the graph of y equals f of x between the points\n

f of x plus h. Let's work out a formula for\n

formula for the average rate of change could\n

A, where a and b are the two locations on\n

a I'm using x instead of B, I'm using x plus\n

as f of x plus h minus f of x over x plus\n

the denominator. Because x plus h minus x,\n

quotient formula f of x plus h minus f of\n

looks like a single entity, but it still represents\n

the difference quotient for this function\n

for the difference quotient. That's an F of\n

compute f of x plus h first, I do this by\n

in the formula for the function. So that's\n

minus x plus h plus three. Notice how I use\n

I need to make sure I shove in the entire\n

to be subtracted, not just the x part, so\n

the parentheses here signal that the entire\n

going to go ahead and simplify a bit right\n

I can go ahead and distribute the negative\n

x squared plus XH plus h x plus h squared.\n

plus 2x H plus two h x plus two h squared\n

are actually the same, I can add them up to\n

I can get that part. Now, I'm going to write\n

going to be this thing right here, minus f\n

in parentheses to make sure I subtract the\n

And now I noticed that a bunch of things cancel\n

add to zero, the minus x and the x add to\nzero

and the three and the minus three, add to\n

minus h. Finally, I'll write out the whole\n

by H. I can simplify this further, because\n

of the numerator. If I factor out this H,\n

h, these two H's cancel, and I'm left with\n

one. This difference quotient will become\n

difference quotient for smaller and smaller\n

and ending up with an expression that has\n

or slope of the function itself. In this video,\n

B minus A to calculate an average rate of\n

h minus f of x over h to calculate and simplify\n

the idea of limits through some graphs and\n

I used to eat at Julia sushi and salad bar\n

food there cost $10 a pound. But if you happen\n

you got a free lunch. So let's let y equals\n

dollars as a function of its weight x in pounds.\n

f of x as a piecewise defined function. It\n

Because there are two situations, the weight\n

be different from one pound. If the weight\n

is zero. But if x is different from a pound,\n

function is going to follow the line y equals\n

is going to have a value of zero and not 10.\n

a place where a point is missing on the graph.\n

then the F of X values that is the y values\n

we say that the limit as x approaches one\n

we can write, as x approaches one, f of x\n

at one is actually equal to zero, not 10.\n

and the value of f at one, are not equal.\n

limit here as x goes to one doesn't care about\n

the value of f, when x is near one, in general,\n

a and L, the limit as x goes to a of f of\n

close to L, as x gets arbitrarily close to\n

of x heads towards L. Let me draw this as\n

the limit as x goes to a of f of x is L. Because\n

a of x values, I can guarantee that my y values,\n

interval around L. Now the limit doesn't care\n

the function, if I change the functions value\n

at a, the limit is still L. But the limit\n

If, for example, the function didn't even\n

could no longer say that the limit, as x approached\n

words, the limit of f of x is L, only if the\n

a from both the left and the right. For the\n

x goes to two of g of x does not exist. Because\n

approach two from the left and the y values\n

from the right. Although the limit doesn't\n

exists. And we write this as limit as x goes\n

one. The superscript minus sign here means\n

In other words, the x values are less than\n

talk about right sided limits. In this example,\n

side of g of x is three. And here, the superscript\n

from the right side. In other words, our x\n

to. In general, the limit as x goes to a minus\n

L, as x approaches a from the left, and the\n

our means that f of x approaches our as x\n

left or from the right are also called one\n

look at the behavior of y equals h of x graph\n

negative two from the right, our Y values\n

any real number we might choose. We can write\n

as x goes to negative two from the right of\n

negative two from the left, the y values are\n

negative real number we might choose. In terms\n

goes to negative two from the left of h of\n

limit of h of x as x goes to negative two,\n

means we have to approach negative two from\n

because the limits from the left and the limit\n

I want to mention that some people say that\n

the left also do not exist. Because the functions\n

to say that these limits do not exist as a\n

and negative infinity. This video introduced\n

infinite limits. This video gives some examples\n

f of x graph below, let's look at the behavior\n

negative one, one, and two. Let's start with\n

negative one from the last, the y value is\n

x approaches negative one from the right,\n

x approaches negative one, and we don't specify\n

only say that the limit does not exist, because\n

Now let's look at the limit as x is approaching\n

we get a limiting y value of two, we approach\n

two. So both of these left and right limits\nare equal to

and therefore the limit as x goes to one of\n

f of one f of one itself does not exist. The\n

x equals one just what happens when x is near\none.

Finally, let's look at the limit as x goes\n

is going to negative infinity. And on the\n

can say the limit as x goes to two is negative\n

as x goes to two does not exist. This is the\n

it carries more information. What values of\n

let's see. Negative one and two are the only\n

limits can fail to exist, we've seen at least\n

where the limit from the left is not equal\n

a where we're calculating the limit at. So\n

seen examples where they're vertical asymptotes.\n

limit fails to exist because of the unbounded\n

to infinity. There's one other way that limits\n

not quite as frequently. And that's wild behavior.\n

Let's look at an example that has this wild\n

the one of the most classic examples as the\n

or sometimes you'll see sine one over x. If\n

and zoom in near x equals zero, you're gonna\n

It just keeps oscillating up and down and\n

towards zero, pi over x is getting bigger\n

these oscillations between one and negative\n

From the other side, when x is negative, you'll\n

faster and faster. As x goes to zero

here, this top values up here at one, and\n

to hit it negative one. Now, when you try\n

zero, well, the y values are going through\n

one and one infinitely often as x goes to\n

limit can settle at. And so the limit as x\n

In this video, we saw three types of examples\n

exist because the one sided limits on the\n

can fail to exist because of vertical asymptotes.\n

wild behavior. And the function fails to settle\n

limit laws, rules for finding the limits of\n

and functions. Let's start with an example.\n

of f of x is 30. And the limit as x goes to\n

x goes to seven of f of x divided by negative\nthree times g of

Wilson's f of x is heading towards 30. And\n

that the quotient should head towards 30 divided\n

in calculating this limit, by plugging in\n

using the limit loss, which are now state.\n

and that the limits as x goes to a of f of\n

is not as limits that are infinity or negative\n

plus g of x is equal to the limit of f of\n

the limit of the sum is the sum of the limits.\n

the difference of the limits. The limit of\n

f of x. And the limit of the product is the\n

is the quotient of the limits, provided that\n

Since we can't divide by zero. Let's use these\n

as x goes to two of x squared plus 3x plus\n

rule about quotients allows us to rewrite\n

limits, provided that the limits of the numerator\n

of the denominator is not zero. But we'll\n

fact hold. Next, we can use the limit rule\n

of limits. And we can rewrite the limit in\n

we can use the limit rule about products to\n

of the limit. That's because x squared is\n

is really the limit of x times x, which by\n

limit of x, which is the limit of x quantity\n

we can now use the limit rule about multiplying\n

times x as three times the limit of x. And\n

we've got things broken down into bite sized\n

Notice that the limit as x goes to two of\n

2x, heads towards two, so we can replace all\n

two. So we get two squared plus three times\n

two of six, well, six doesn't have any x's\n

at six, and the limit here is just six. So\n

set limit of six with six. And on the denominator\n

arithmetic, this simplifies to 16/11. Notice\n

faster, just by substituting in the value\n

fact, that's the beauty of the lemon laws,\n

functions, just by plugging in the number\n

in that number doesn't make the denominator\n

in the value does make the done at zero. And\n

techniques for handling the situation. In\n

It's important to note, these limit laws only\n

actually exist as finite numbers.

If the limit of one or both of the component\n

just simply don't apply. And instead, we have\n

the limit of a sum, difference product or\n

theorem, which is another method for finding\n

we have a function g of x. We don't know much\n

is near one, g of x is greater than or equal\n

in red, and less than or equal to 3x squared\n

on the picture, G has to lie between the red\n

it could look something like this. What can\n

to one? Well, if you notice, the red curve\n

of four as x goes to one. And since the green\n

blue curve, its limit must also be for this\n

Now, let's say the squeeze theorem in general,\n

x g of x and h of x. And let's suppose that\n

is less than or equal to h of x, at least\n

doesn't necessarily have to hold for x equal\n

limits. And limits don't care what happens\n

Let's suppose also, that like in the previous\n

same limit as x approaches a. So we're going\n

f of x is equal to the limit as x goes to\n

The picture looks a lot like the previous\n

what happens at x equals a, for example, g\n

could be for example, way up here. Since g\n

h of x, which both have the same limit l at\n

to a of g of x is equal to l also. And that's\n

theorem, and the sandwich theorem, three very\n

geovax being trapped here, in between lower\n

there. To find the limit as x goes to zero\n

remember that sine one over x by itself has\n

the limit as x goes to zero of sine one over\n

settles down to a single finite value. At\n

x squared sine of one over x also wouldn't\n

the product rule and say that the limit of\n

But in fact, the product rule only applies\n

since the second limit doesn't exist, the\n

whether the limit that we're interested in\n

rule. But it turns out, we can use the squeeze\n

than the first example, because in the first\n

bounding functions should be. And in this\n

look at a graph of x squared sine one over\n

Let's use algebra to see what those two bounding\n

of one over x is always between one and negative\n

one and negative one. And if we multiply this\n

x squared is less than or equal to x squared,\n

to x squared. Notice that x squared is always\n

flipping any of the inequality signs when\n

squared and minus x squared are good bounding\n

as x goes to zero of x squared is zero, and\n

squared is also zero, we can conclude by the\n

to zero of x squared, sine of one over x is\n

these two functions with the same limit. the\n

limits when you happen to have a function\n

two other functions with the same limit. The\n

where we have a crazy oscillating trig function,\n

of our bounding functions. In this video,\n

tricks. All these limits are of the zero over\n

this means that the limits are of the form\n

limit of f of x equals zero and the limit\n

let's look at the limit as x goes to one of\n

Notice that the numerator and the denominator\n

calculate this limit, we want to simplify\n

is to factor it. So let's rewrite this as\n

times x squared plus x plus one. We're factoring\n

We can also factor the denominator as a difference\n

Now, as long as x is not equal to one, we\n

one. And so the limit of this expression is\n

Now we can just plug in one, because plugging\n

denominator of two. And that way, we've evaluated\n

of five minus z quantity squared minus 25.\n

in x equals zero in the numerator, we get\n

This time instead of factoring, the trick\n

this limit as the limit as t goes to zero

25 minus 10 z plus z squared minus 25. I just\n

by z. Since 25 minus 25 is zero, I just have\n

Now what? Well, I could factor out the Z here.\n

original limit is the same as this limit.\n

10 as my answer. This third example is also\n

zero, and the denominator is also going to\n

the expression by adding together the fractions\n

which is r plus three, times three. So rewriting,\n

I multiply that by three over three, in order\n

minus 1/3, which I have to multiply by R plus\n

continuing to rewrite, I have in the numerator,\n

the denominator of R plus three times three,\n

then this entire fraction is still divided\n

three minus r minus three divided by R plus\n

minus three is zero. So I can rewrite this\n

And now instead of dividing by r, which is\n

one over R. R divided by R is one. So this\n

r plus three, times three. So finally, I'm\n

go ahead and let our go to zero. And by plugging\n

one over zero plus three times three, or negative\n

because involve square roots can be hard to\n

dealing with square roots that works here,\n

the expression that we're given, and multiply\n

case, because the numerator is the place where\n

B, I just mean a plus b conjugate of A plus\n

on the numerator, by something, I also have\n

so that I won't alter the value of the expression.\n

the expression down here that looks more complicated,\n

clear up and become simpler. So Multiplying\n

plus two times the square root of x plus three\n

three minus four. On the denominator, I get\n

2x minus the square root of x plus three minus\n

square root of x plus three squared is just\n

and this expression are opposites, so they\n

I just have x plus three, and then I still\n

looks a little messy. I'll just copy it over\n

x minus one. Notice that when x goes to one,\n

in fact, as I let x go to one, that denominator,\n

cancel out to zero. So I still got a zero\n

can use one of the previous tricks of factoring.\n

of these two expressions as a square root\n

and I could maybe factor a two out of these\n

So we have x squared of x plus three times\n

then I have a plus two times x minus one.\n

promising. So now I've got an x minus one\n

one from each of these two expressions, I'm\n

times the square root of x plus three, plus\n

to one, I can cancel those. And my limit simplifies\n

plus two, plugging in one, I get a one on\n

is two plus two, four on the denominator,\n

here, another zero over zero indeterminate\n

going to want to take cases, because the absolute\n

the cases, if X plus five is greater than\n

five, then the absolute value of this positive\nnumber is just

On the other hand, if X plus five is less\n

negative five, than the absolute value of\n

make the expression x plus five, turn it into\n

front. Now let's look at one sided limits.\n

a situation where x is less than negative\n

can rewrite the absolute value by taking the\n

a zero over zero and determinant form. But\n

factor out a two cancel, we got the limit\n

We do the same exercise on the right side.\n

than negative five. So we're in this situation\n

value with the stuff inside and again factoring\n

So we have a left limit, and a right limit\n

the limit does not exist. So we've seen a\n

zero or zero indeterminate form. And we've\n

For the first example, we use factoring. For\n

we multiplied out. For the third example,\n

to simplify things. The next example uses\n

the last example, we used cases and looked\nat one sided limits.

The limit law about quotients tells us that\n

of the limit, provided that the limits of\n

that the limit of the function on the denominator\n

the limit of the function on the denominator\n

answer that question. In fact, there are two\n

It could be that even though the limit on\n

the limit on the numerator exists and is not\n

limits are zero. We'll focus on the first\n

And we'll look at the second situation later\non.

In this example, the limit of the numerator,\n

can see by plugging in three for x. But the\n

is zero. So we're exactly in one of these\n

nonzero number, but the denominator goes to\n

three from the left first. As we approach\n

that are slightly less than three numbers\n

in those numbers into the expression, here\n

116 1196, and 11,996. If even without a calculator,\n

closely by just thinking about the fact that\n

is about negative four times three, so about\n

is negative 0.1. That quotient of two negative\n

Similarly, we could approximate the value\n

divided by 0.01, which is 1200. And approximate\n

do it exact answers on our calculator or approximations\n

are positive numbers that are getting larger\n

left. This makes sense. Because if we look\n

from the left, the numerator is getting close\n

And the denominator, since x is less than\n

negative over negative is a positive. And\n

denominators are getting smaller and therefore\n

in magnitude. So we can conclude that our\n

similar argument By looking at the limit,\n

it's supposed to be an X minus three here.\n

bigger than three 3.1 3.01 3.01. And again,\n

figure out the answers are negative 124. Negative\n

approximating argument. This answer is approximately\n

is negative 120, and so on. Like before, if\n

denominator, we can see that as x goes to\n

But our denominator is a positive number,\n

where x is bigger than three, and therefore,\n

getting bigger and bigger in magnitude as\n

is still getting tinier and tinier, while\n

12. So in this case, we're getting a negative\n

so that makes a limit of negative infinity.\n

and our limit on the right is negative infinity,\n

is that it does not exist. Now let's look\n

negative four of 5x, over the absolute value\n

numerator is just negative 20 by plugging\n

denominator is zero. Because we've got an\n

screaming out at us to look at cases. Remember\n

going to equal just x plus four, if x plus\n

greater than negative four, however, if x\n

x plus four will be negative. So taking the\n

to make a negative expression positive. Alright,\n

at the limit as x goes to negative four, from\n

negative four from the left, x is going to\n

to be in this situation here, where the absolute\n

that the limit as x goes to negative four\n

limit of 5x over negative x plus four. Now\n

four from the left, the numerator here is\n

20. The denominator, since x is less than\n

of it is positive. So our quotient is negative.\n

while the numerators will pretty level at\n

bigger and bigger magnitude a limit of negative\n

goes to negative four from the right, in this\n

four. So we're in this case where the absolute\n

now the numerator is still gonna be a negative\n

bigger than negative four, slightly to the\n

number. Negative or over a positive is a negative.\n

tiny, the fractions getting huge in magnitude.\n

in this case, look at what's going on, we've\n

and a negative infinity limit on the right.\n

to negative four of 5x, over the absolute\n

infinity. In fact, we can confirm that by\n

the graph near x equals negative four, it's\n

asymptote at x equals negative four, like\n

of f of x is equal to something that's not\n

zero, then the limit of the quotient could\n

example. It could also be infinity, or it\n

not exist, if the one sided limits are infinity\n

other. Now, what about this second situation\n

of f of x is zero, and the limit of f of g\n

of the quotient in this situation? Well, in\n

quotient, it could exist and be any finite\n

it could not exist at all. In fact, in this\n

or zero indeterminate form, anything could\n

but in some ways, the most fun situation of\n

techniques for dealing with 00 indeterminate\n

techniques to evaluate these these mysterious\n

the limits of quotients, when the limit of\n

when the limit of the numerator was not zero,\n

And we saw that these situations corresponded\n

for the limit of the quotient of infinity,\n

on one side and negative infinity on the other.\n

we look at the limits in this situation when\n

heading towards zero. And when anything can\n

of lines. Here we're given the graph of a\n

format for the equation of a line is y equals\n

B represents the y intercept, the y value,\n

is equal to the rise over the run. Or sometimes\n

over the change in x values. Or in other words,\n

where x one y one and x two y two are points\non the line.

While we could use any two points on the line,\n

points where the x and y coordinates are integers.\n

grid points. So here would be one convenient\n

point to use. The coordinates of the first\n

is let's say, five negative one. Now I can\n

the run. So as I go through a run of this\n

especially gonna be a negative rise or a fall\n

see counting off squares, this is a run of\n

three, so my slope is going to be negative\n

squares. But I could have also gotten it by\n

the difference of my x values. That is, I\n

from my difference in Y values, and divide\n

five minus one, that gives me negative three\n

three fourths. Now I need to figure out the\n

just read it off the graph, it looks like\n

accurate, I can again use a point that has\n

coordinates. So either this point or that\n

off with my equation y equals mx plus b, that\n

b. And I can plug in the point one, two, for\n

three fourths times one plus b, solving for\n

fourths plus b. So add three fourths to both\n

b. So b is eight fourths plus three fourths,\n

what I eyeballed it today. So now I can write\n

negative three fourths x plus 11 fourths by\n

the equation for this horizontal line. a horizontal\n

as y equals mx plus b, m is going to be zero.\n

some constant. So if we can figure out what\n

it's to let's see, this three, three and a\n

directly, y equals 3.5. For a vertical line,\n

I mean, if you tried to do the rise over the\n

be divided by zero and get an infinite slope.\n

equation of the form x equals something. And\n

that all of the points on our line have the\n

y coordinate can be anything. So this is how\n

In this example, we're not shown a graph of\n

through two points. But knowing that I go\n

for the line. First, we can find the slope\n

the difference in x values. So that's negative\n

is negative five thirds. So we can use the\n

And we can plug in negative five thirds. And\n

still get the same final answer. So let's\n

negative five thirds times one plus b. And\n

thirds plus five thirds, which is 11 thirds.\n

thirds x plus 11 thirds. This is method one.\n

of the equation. It's called the point slope\n

to m times x minus x naught where x naught\n

is the slope. So we calculate the slope the\n

over a difference in x values. But then we\n

the point one, two will work, we can plug\n

this point slope form, that gives us y minus\n

one. Notice that these two equations, while\n

Because if I distribute the negative five\n

So we've seen two ways of finding the equation\n

and using the point slope form. In this video,\n

a line if you know the slope. And you know\n

for the line if you know two points, because\n

and then plug in one of those points. To figure\n

forms for the equation of a line the slope\n

is the slope, and B is the y intercept. And\n

m times x minus x naught, where m again is\n

on the line. This video is about rational\n

rational function is a function that can be\n

Here's an example. The simpler function, f\n

a rational function, you can think of one\n

of this rational function is shown here. This\n

polynomial. For one thing, its end behavior\n

is the way the graph looks, when x goes through\n

numbers, we've seen that the end behavior\n

cases. That is why marches off to infinity\n

big or really negative. But this rational\n

Notice, as x gets really big, the y values\n

And similarly, as x values get really negative,\n

equals three, I'll draw that line, y equals\n

asymptote. A horizontal asymptote is a horizontal\n

to as x goes to infinity, or as X goes to\n

else that's different about this graph from\n

x gets close to negative five. As we approach\n

our Y values are going down towards negative\n

negative five from the left, our Y values\n

say that this graph has a vertical asymptote\n

is a vertical line that the graph gets closer\n

really weird going on at x equals two, there's\n

at x equals two is dug out. That's called\n

of the graph where the function doesn't exist.\n

of our rational functions graph, I want to\n

have predicted those features just by looking\n

We need to look at what our function is doing\n

really big negative numbers. Looking at our\n

to be dominated by the 3x squared term when\n

x squared is going to be absolutely enormous\n

positive or negative number in the denominator,\n

squared term. Again, if x is a really big\n

a million squared will be much, much bigger\n

For that reason, to find the end behavior,\n

we just need to look at the terms on the numerator\n

the highest exponent, those are the ones that\n

really big, our functions y values are going\n

which is three. That's why we have a horizontal\n

three. Now our vertical asymptotes, those\n

function is zero. That's because the function\n

And when we get close to that place where\n

dividing by tiny, tiny numbers, which will\n

So to check where our denominators zero, let's\n

go ahead and factor the numerator and the\n

see, pull out the three, I get x squared minus\n

into X plus five times x minus two, I can\n

that's three times x minus two times x plus\n

is equal to negative five, my denominator\n

zero. That's what gives me the vertical asymptote\n

x equals two, the denominators zero, but the\n

the x minus two factor from the numerator\n

my function that agrees with my original function,\n

when x equals two, the simplified function\n

it's zero over zero, it's undefined. But for\n

x equals two, our original function is just\n

our function only has a vertical asymptote\n

two, because the x minus two factor is no\n

it does have a hole at x equals two, because\n

even though the simplified version is if we\n

just plug in x equals two into our simplified\n

of three times two plus two over two plus\n

thirds. So our whole is that to four thirds.\n

detail, let's summarize our findings. We find\n

where the denominator is zero. The holes happen\n

zero and those factors cancel out. The vertical\n

denominator is zero, we find the horizontal\n

term on the numerator and the denominator,\n

three examples. In the first example, if we\n

to 5x over 3x squared, which is five over\n

is going to be huge. So I'm going to be dividing\n

to be going very close to zero, and therefore\n

zero. In the second example, the highest power\n

two thirds. So as x gets really big, we're\n

we have a horizontal asymptote at y equals\n

power terms x squared over 2x simplifies to\n

is getting really big. And therefore, we don't\n

going to infinity, when x gets through goes\n

to negative infinity when x goes through a\n

end behavior is kind of like that of a polynomial,\n

In general, when the degree of the numerator\n

we're in this first case where the denominator\n

and we go to zero. In the second case, where\n

of the denominator equal, things cancel out,\n

y value, that's equal to the ratio of the\n

case, when the degree of the numerator is\n

then the numerator is getting really big compared\n

asymptote. Finally, let's apply all these\n

the video and take a moment to find the vertical\n

for this rational function. To find the vertical\n

the denominator is zero. In fact, it's going\n

the denominator. Since there if there are\n

of a vertical asymptote. The numerator is\n

times x plus one for the denominator, first\n

more using a guess and check method. I know\n

to the to x squared, and I'll need a three\n

one. Let's see if that works. If I multiply\n

does give me back my 2x squared plus 5x minus\n

that I have a common factor of x in both the\n

me I'm going to have a hole at x equals zero.\n

by cancelling out that common factor, and\n

to zero. So the y value of my

is what I get when I plug zero into my simplified\n

one over two times zero minus one times zero\n

or minus one. So my whole is at zero minus\n

that make my denominator zero will get me\n

asymptote, when 2x minus one times x plus\n

one is zero, or x plus three is zero. In other\n

three. Finally, to find my horizontal asymptotes,\n

term in the numerator and the denominator.\n

bottom heavy, right? When x gets really big,\n

means that we have a horizontal asymptote\n

of our graph, the whole, the vertical asymptotes\n

would give us a framework for what the graph\n

at y equals zero, vertical asymptotes at x\n

at a hole at the point zero minus one. plotting\n

of graphing program, we can see that our actual\n

that the x intercept when x is negative one\n

rational function or reduced rational function\n

the numerator that doesn't make the denominator\n

X intercept is where the y value of the whole\n

how to find horizontal asymptotes of rational\n

terms, we learned to find the vertical asymptotes\n

of the functions, the holes correspond to\n

zero, his corresponding factors cancel. The\n

that make the denominator zero, even after\n

in the numerator denominator. This video focuses\n

x goes through arbitrarily large positive\n

on these ideas before in the past when you\n

video, we'll talk about the same ideas in\n

what happens to the function f of x drawn\n

numbers? Well, the arrow here on the end is\n

x gets bigger and bigger, the values of y,\n

we can write this in the language of limits\n

of f of x is equal to one. Now what happens\n

and larger negative numbers, by larger and\n

are negative but are larger and larger in\n

negative 10, negative 100, and negative a\n

continues, it looks like f of x, even though\n

value of two. So we say that the limit as\n

to limits in which x goes to infinity, or\n

The phrase limits at infinity should be contrasted\n

limit means that the y values, or the F of\n

limits and infinity correspond to horizontal\n

limits correspond to vertical asymptotes.\n

a horizontal or vertical asymptote is when\n

same time, we'll see an example of this on\n

of infinity for these two functions, g of\n

the function e to the minus x, and it has\n

at y equals zero. So the limit as x goes to\n

head to the left, and x goes through larger\n

settle down to a particular finite value,\n

that the limit as x goes to minus infinity\n

look at the graph of y equals h of x. Please\n

out the limits of infinity for this function.\n

is negative infinity. Because as x goes to\n

than any finite number. Now as X goes to negative\n

settle down at a particular number. So we\n

of h of x does not exist. Finally, let's look\n

looking at their graphs first, to find the\n

Let's think about what happens to one over\n

numbers. As x gets bigger and bigger, one\n

as x goes to infinity of one over x equals\n

goes to negative infinity, let's look at what\n

are larger and larger in magnitude. Now one\n

but they're still getting smaller and smaller.\n

infinity of one over x is also zero. We can\n

x goes to infinity of one over x cubed. As\n

So one of our x cubed has to go to zero. To\n

over the square root of x, notice that as\n

goes to infinity. So one over the square root\n

of these limits are equal to zero. Both of\n

because both have the form of the limit as\n

where R is a number greater than zero. In\n

really x to the R, where R is one half. In\n

over x to the R is always equal to zero. Whenever\n

same thing about the limit as x goes to negative\n

we avoid exponents, like one half that don't\n

values of r, as x goes to negative infinity,\n

magnitude. And so one over x to the R is getting\n

towards zero. Notice that this is no longer\n

something like r equals negative two, because\n

And the limit as x goes to infinity of x squared\n

In this video, we looked at examples of limits\n

infinity. And we saw that those limits could\n

infinity, or not exist. This video gives some\n

at infinity of rational functions. Let's find\n

function. The numerator and the denominator\n

arbitrarily large as x goes to infinity. One\n

the graph of the numerator looks like a parabola\n

looks like some kind of cubic. So something\n

goes to infinity, y also goes to infinity.\n

form. And just like the zero have over zero\n

over infinity and determinant form could turn\n

to use algebra to rewrite this expression\n

evaluate. Specifically, we're going to factor\n

from the numerator, and then from the denominator.\n

the highest power I see in the numerator,\n

out of 5x squared, I get five, when I factor\n

dividing negative 4x by x squared, so I get\n

this works by distributing the x squared and\n

Now the highest power of x SC, and the denominator\n

for each from each of those terms, I get a\n

Because factoring out an x cubed is the same\n

writing the x cubed on the side. Now, we can\n

the top and the bottom to get the limit of\n

two minus 11 over x plus 12 over x squared.\n

to zero, because I'm dividing for by larger\n

to zero, and 12 over x squared goes to zero.\n

going to zero times something that's going\n

zero times five halves, which is just zero.\n

do these last steps, which is fine, because\n

Something that wasn't true from my original\n

In this example, we're asked to find the limit\n

rational expression. I encourage you to stop\n

this example, the highest power of x in the\n

in the denominator is also x cubed. Factoring\n

x cubed times three plus six over x plus 10\n

factoring out the x cubed from the denominator,\n

plus five over x cubed. Now the x cubes cancel,\n

dust clears here, our limit is just three\n

power in the numerator is x to the fourth,\n

x squared. So we factor out the x to the fourth\n

the denominator and cancel as much as we can.\n

So our limit is the same as the limit of x\n

goes to negative infinity, x squared is positive\n

by negative fifth turns it negative, but doesn't\n

arbitrarily large. Therefore, our final limit\n

same three examples again, more informally,\n

In the first example, the term 5x squared\n

is much larger than x when x is large. In\n

x cubed dominates, because x cubed is much\n

If we ignore all the other terms in the numerator\n

terms, which have the highest powers, then\n

squared over 2x cubed, which is the same as\n

2x, just by canceling Xs, which is zero as\n

focus on the highest power terms in the numerator\n

get the limit of 3x cubed over 2x cubed, which\n

is just three halves. In the third example,\n

and negative 5x squared. And we rewrite the\n

and simplify, and we get the limit as x goes\n

five, which is negative infinity as before,\n

at the highest power terms, lets you reliably\n

infinity when the degree of the numerator\n

then the limit as x goes to infinity, or negative\n

when the degree of the numerator is equal\n

limit is just the quotient of the highest\n

as the limit in the second example. And finally,\n

than the degree of the denominator, then the\n

Like it was in the third example. These shortcut\n

to also understand the technique of factoring\n

can be used more generally. This video gave\n

of rational functions. First, there's the\n

terms and simplifying. Second, there's the\n

the numerator and the degree of the denominator,\n

the past, you may have heard an informal definition\n

continuous, if you can draw it without ever\n

develop a more precise definition of continuity\n

at some examples of functions that are discontinuous,\n

In order to better understand what it means\n

and try to draw graphs of at least two different\n

ways. One common kind of discontinuity is\n

a jump discontinuity, if its graph separates\n

This particular function can be described\n

equations, f of x equals to x when x is less\n

x plus two, when x is greater than one. Another\n

You may have encountered these before, when\n

holes in them, for example, the function f\n

minus four divided by x minus four, which\n

looks like the graph of x minus three squared.\n

because you could get rid of it by plugging\n

value when x equals four. So in this case,\n

not equal to four, but you'd want it to have\n

would amount to plugging the hole and making\n

function had a removable discontinuity because\n

a function could also have a removable discontinuity,\n

x equals four, for example, too high or too\n

can also occur at a vertical asymptote, where\n

example, the rational function g of x is one\n

at x equals two. Occasionally, you may encounter\n

of these. For example, the graph of the function\n

at x equals zero, because of the wild oscillating\n

at x equals a, we need it to avoid all of\n

we can insist that the functions limit has\n

or removable discontinuity, we can insist\n

avoid the other kind of removable discontinuity\n

wrong place. We can insist that the limit\n

a. Sometimes the definition of continuity\n

and the first two conditions are implied.\n

exclude jump this continuity and removal this\n

and wild discontinuities. For example, in\n

at x equals two, because it fails to have\n

a value at x equals two. In our wild, this\n

exist at x equals zero, so the function can't\n

where this function f is not in us and why,\n

yourself. The functions not continuous at\n

not defined there, the functions not continuous\n

In the language of limits, we say that the\n

two, the limit of the function exists and\n

is down here at negative one. So the function\n

equal the value at x equals three, the function\n

limit doesn't exist. Notice that x equals\n

a corner, the function still continuous. Because\n

of the function is also two. The function\n

two, because the limit doesn't exist at x\n

is one, while the limit from the right is\n

function at x equals negative two is equal\n

side. That is, f of negative two is equal\n

Notice that we can't say the same thing about\n

is zero, while the value of the function is\n

we say that f is continuous from the left,\n

at x equals one, the function is not continuous,\n

the limit from the right is equal to the value\n

from the left here, because the limit from\n

value of the function is one. In general,\n

the left at x equals j. If the limit as x\n

to f of a, and a function is continuous from\n

goes to a from the right of f of x equals\n

continuous from the left if the endpoint is\n

is continuous from the right, if the endpoint\n

gave a precise definition of continuity at\n

is continuous at the point x equals a. If\n

equal to the functions value

In a previous video, we gave a definition\n

we'll discuss continuity on an interval and\n

f of x is continuous on the open interval\n

in that interval. For x to be continuous on\n

continuous on every point in the interior\n

from the right at B. And from the left at\n

on half open intervals. For example, on half\n

closed at sea, or the other way around, or\n

and so on. In all of these cases, we require\n

interval, and left or right continuous on\n

So on what intervals is this function geovax\n

part, the arrows indicate it keeps on going.\n

to negative one, not including the endpoint\n

we can include the endpoint this time. So\n

again, on this last section, we can't include\n

defined there. So what kinds of functions\n

continuous everywhere? And by everywhere,\n

infinity? Well, polynomials are a great example.\n

of x is another common example. There are\n

on the whole real line. I'll let you see if\n

if we ask the second question, what kinds\n

we get a lot more answers, not only polynomials,\n

f of x equals 5x minus two over x minus three\n

of a rational function, even though it's not\n

when x equals three or negative four, it is\n

and four are not in the domain of this rational\n

inverse trig functions, log and either the\n

we normally encounter are continuous on their\n

the whole real line. For example, for natural\n

and that's where the function is continuous.\n

and quotients of continuous functions are\n

y equals sine of x plus the natural log of\n

of continuous functions are continuous on\n

y equals ln of sine of x is continuous, where\n

intervals where sine is positive. Since continuity\n

possible to use our knowledge of which functions\n

if we want to find the limit as x goes to\n

we can evaluate this limit just by plugging\n

We're using the definition of continuity here\n

to the value of the function. The second example\n

inside is not continuous at x equals two.\n

But as X approaches 2x squared minus four\n

as x plus two times x minus two over two times\n

as x plus two over two pi. For x not equal\n

here, approaches two plus two over two times\n

the limit as x goes to two of x squared minus\n

And therefore, the limit as x goes to two\n

of two pi, which is again equal to one. We're\n

and a property of continuous functions, which\n

g of x is equal to f of the limit as x goes\n

In other words, for continuous functions,\n

That's all for continuity on intervals and\n

theorem says that if f is a continuous function,\n

number, in between F of A and F of Bay, n\n

In other words, if n is a number between F\n

c in the interval a, b, such that f of c equals\n

values for C, it could be right here, since\n

here, or here, I'll just mark the middle one.\n

applied to continuous functions. If the function\n

and, and never achieve that value. When application\n

the existence of roots or zeros of equations,\n

such that P of C is zero, we're going to want\n

with n equal to zero. Our polynomial is defined\n

But the trick here is to pick an interval\n

B is positive, or vice versa. So that the\n

P has to pass through zero in between. I'm\n

calculate a few values of p. So P of zero\n

P of one is going to be five minus three minus\n

So in this very lucky example, the first two\n

B, so we can just let A be equal 01. Because\n

is a negative number. So actually, the graph\n

looks more like this. But in any case, by\n

to be a number c, in between, in this case,\n

this intermediate value of zero. And that\n

we know it somewhere in the interval zero\n

for our polynomial. There may be other real\n

one. The intermediate value theorem has lots\n

For example, suppose you have a wall that\n

height of the wall varies continuously as\n

intermediate value theorem can be used to\n

opposite places on the wall with exactly the\n

to show this in this video, we stated the\n

continuous functions, and talked about a couple\n

trig functions, sine, cosine, tangent, secant,\n

For a right triangle with sides of length\n

sine of theta as the length of the opposite\n

opposite to our angle theta has measure a\n

measure C. So that would be a oversee for\n

as the length of the adjacent side over the\n

the side adjacent to theta. Of course, the\n

it's special as the high partners so we don't\n

would be B oversea. tangent of theta is the\n

length. So that would be a over b. The pneumonic\n

is opposite over hypotenuse. Cosine is adjacent\n

adjacent. In fact, there's a relationship\n

tangent of theta is equal to sine of theta\n

that's, that's because sine of theta over\n

opposite over hypotenuse, divided by cosine,\n

these fractions by flipping and multiplying

the high partners, length cancels, and we\n

by definition, tangent of theta. There are\n

in terms of sine, cosine and tangent. First\n

that's one over cosine of theta. So it's going\n

which is the high partners over the adjacent,\n

of theta is defined as one over sine theta.\n

which is the high partners over the opposite.\n

C over a. Finally, cotangent of theta is defined\n

one over opposite over adjacent, flip and\n

which in this case is b over a. So notice\n

secant are the reciprocals of the values for\n

use these definitions to find the exact values\n

in this triangle. I'll start with sine of\n

Well, for this angle theta, the opposite side\n

partners has measured five. So sine theta\n

over hypotenuse, but I don't know the value\n

find it using the Fagor in theorem says I\n

squared, I'll call this side length A plus\n

triangle is equal to c squared, where c is\n

two squared equals five squared, which means\n

is 21. So A is plus or minus the square root\n

of a side of a triangle, I can just use the\n

of cosine theta, I can write it as adjacent,\n

News, which is five, tangent theta is the\n

to be two over the square root of 21. To compute\n

so that's going to be one over square root\n

root of 21. The reciprocal of my cosine value.\n

going to be the reciprocal of my sine, so\n

a tan theta, so it's going to be the reciprocal\n

Finally, we'll do an application. So if we\n

as the angle from the horizontal bar of 75\n

we want to find out how high it is. I'll call\n

known quantities this angle and this hypothesis\n

is the opposite side of our triangle. So if\n

hypothesis, then we can relate these known\n

amount y, which is the opposite and are known\n

gives that y is 100 meters times sine of 75\n

sine of 75 degrees, be sure you use degree\n

the 75. When I do the computation, I get a\n

places. Notice that we're ignoring the height\n

To remember the definitions of the trig functions,\n

fact that secant is the reciprocal of cosine\n

the reciprocal of tangent. In this video,\n

cosine of a 30 degree angle, a 45 degree angle\n

the sine of a 45 degree angle is to use a\n

particular right triangle has I have partners\n

triangle have to add up to 180 degrees, and\n

the remaining angle must also be 45 degrees.\n

sides the same length, I'll call that side\n

use this 45 degree angle here, then sine is\n

out how long this side length is, I'll be\n

Pythagorean theorem says this side length\n

I have hotness squared. So we have that a\n

All right, that is two A squared equals one.\n

minus the square root of one half. Since we're\n

I can just use the positive square root it's\n

of one over the square root of two, which\n

rationalize the denominator by multiplying\n

of two. That gives me a square root of two\n

squared in the denominator, which is the square\n

are the square root of two over two. Now I\n

computing the opposite over the hypotenuse.\n

is square root of two over two hypothesis\n

root of two over two. Cosine of 45 degrees\n

length over this hypothesis. So that's the\n

what would happen if instead of using this\n

triangle, also a 4545 90 triangle with hypotony\n

pause the video for a moment and repeat the\n

I'll call the side length B. Pythagoras theorem\n

squared. So two, b squared equals 25. And\n

be the plus or minus the square root of 25\n

version. And so B is the square root of 25\n

over the square root of two, rationalizing\n

two. Now, sine of my 45 degree angle is opposite\n

five square root of two over two divided by\n

two over two as before, and a similar computation\n

root of two over two as before. This makes\n

ratios of sides. And since these two triangles\n

ratios of sides. To find the sine and cosine\n

triangle with hypotony is one. If we double\n

so a total angle of 60 degrees here, and this\n

6060 triangle, that's an equilateral triangle,\n

side length has length one, this side length\n

which means this short side of our original\n

my original triangle, let's use the Pythagorean\n

x. But agrin theorem says x squared plus one\n

plus a fourth equals 1x squared is three fourths.\n

of three fourths is the positive version,\n

square root of four, which is the square root\n

triangle, again, let's compute the sine of\n

of 30 degrees is opposite of our hypothesis,\n

the hypothesis is one. So we get a sine of\n

of 30 degrees is adjacent over hypothesis.\n

divided by one. To find sine of 60 degrees\n

use this same green triangle and just focus\n

So sine of 60 degrees. Opposite overhype hotness,\n

the square root of three over two. Cosine\n

us one half. I'll summarize the results in\n

angle corresponds to pi over six radians since\n

Similarly, 45 degrees corresponds to pi over\n

pi over three radians. I recommend that you\n

two over two, and root three over two. And\n

two go together. And root two over two goes\n

hard to reconstruct the triangles, you know\n

So it must have the side lengths were the\n

triangle has one side length smaller than\n

half and the larger side must be root three\n

and so root three over two is bigger than\n

fill in the angles, the smaller angle must\n

must be the 60 degree one. In this video,\n

angles 30 degrees, 45 degrees, and 60 degrees.\n

of points on the unit circle. a unit circle\n

Up to now we defined sine and cosine and tangent\n

to find sine of 14 degrees in theory, you\n

of 14 degrees and then calculate the sign\n

length of pi partners. But if we use this\n

things go horribly wrong. When we draw this\n

no way to complete this picture to get a right\n

unit circle, that is a circle of radius one.\n

and a unit circle are related. If you draw\n

one with larger and larger angles, then the\n

Let's look at this relationship in more detail.\n

inside a unit circle, the high partners of\n

which is one, one vertex of the right triangle\n

triangle is at the edge of the circle, I'm\n

A, B. Now the base of this right triangle\n

of the right triangle is B, the y coordinate.\n

sine and cosine of theta, this right here\n

adjacent over hypotenuse, so that's a over\n

x coordinate of this point on the unit circle.\n

that down. For sine of theta, if I use the\n

overhead partners, so B over one, which is\n

of this point in the on the unit circle at\n

the right side triangle definition, its opposite\n

of that as the y coordinate of the point over\n

theta, that can't be part of a right triangle,\n

90 degrees, like now I'll call this angle\n

and y coordinates to calculate the sine and\n

on the end of this line at angle theta, if\n

cosine theta, I'm still going to define as\n

as the y coordinate, and tangent theta as\n

When we use this unit circle definition, we\n

x axis and going counterclockwise. Let's use\n

cosine and tangent of this angle fee. In our\n

are supposed to represent the x&y coordinates\n

of this line segment, that lies at angle fee\n

to the y coordinate cosine fee is equal to\n

by the ratio of the two, which works out to\n

This video gives a method for calculating\n

circle. Starting from the positive x axis,\n

You look at the coordinates of the point on\n

cosine of that angle theta is the x coordinate,\n

of theta is the ratio. This video gives three\n

that can be deduced from the unit circle definition.\n

sine and cosine for angle theta is that cosine\n

is the y coordinate for the point on the unit\n

is what I call the periodic property. This\n

periodic with period two pi. And what that\n

plus two pi, you get the same thing as just\n

write this down, we're assuming that theta\n

theta in degrees, the similar statement is\n

to cosine of theta, we can make the same statements\n

equal to sine of the original angle, here,\n

want to measure the angle in degrees, the\n

equal to sine of theta, we can see why this\n

sine and cosine. This is our angle theta,\n

a full turn around the unit circle to our\n

and theta plus two pi are just two different\n

And since sine and cosine give you the y and\n

they have to have the same value. Similarly,\n

theta minus two pi, the minus two pi means\n

circle clockwise, we still end up in the same\n

two pi, the x coordinate of that position\n

of theta minus two pi is the same thing as\n

we add or subtract multiples of two pi. For\n

the same thing as cosine of theta. This time,\n

and still gotten back to the same place. So\n

the same thing as cosine of pi plus four pi,\n

about the unit circle, pi is halfway around\n

x coordinate of this point right here. Well,\n

so cosine of pi must be negative one. If I\n

well, that's sine of negative 360 degrees,\n

as sine of minus 60 degrees. Thinking about\n

start at the positive x axis and go clockwise\n

And so that's one of the special angles that\n

of negative root three over two. And therefore\n

over two the y coordinate. The next property\n

cosine is an even function, which means that\n

as cosine of theta, while sine is an odd function,\n

the negative of sine of theta. To see why\n

And the angle negative theta. A negative angle\n

direction from the positive x axis.

The coordinates of this point by definition,\n

of this point are cosine negative theta, sine\n

two points have the exact same x coordinate,\n

of negative theta, while their y coordinates\n

This one's positive and this one's negative.\n

the negative of sine of theta. Let's figure\n

Well, we know that tan of negative theta,\n

Well, we know that sine of negative theta\n

cosine of negative theta is cosine of theta.\n

over cosine theta, which is negative tan of\n

negative of tan of theta, tan theta is an\n

is the Pythagorean property, which says that\n

squared is equal to one. A lot of times this\n

cosine squared theta plus sine squared theta\n

theta just means you take cosine of theta\n

Pythagorean property, because it comes from\n

triangle on the unit circle. I'll call this\n

here are cosine theta sine theta. Since this\n

of my right triangle has length one, the base\n

same thing as the x coordinate of this point.\n

of the point sine theta. Now the Pythagorean\n

plus that so that squared equals one squared,\n

that gives me the Pythagorean property. But\n

of cosine given values of sine and vice versa.\n

t is negative two sevenths. And T is an angle\n

angle lies in quadrant three, that means the\n

three. One way to find cosine of t is to use\n

t is equal to one. That is cosine of t squared\n

I can write this as cosine of t squared plus\n

is equal to one minus 4/49, which is 4540\n

that goes for the cosine t is plus or minus\n

or minus the square root of 45 over seven.\n

know that cosine of t, which represents the\n

Therefore, cosine of t is going to be negative\n

to solve this problem using the Pythagorean\n

look at the fact that sine of t is negative\n

for now, we can think of this information\n

angle theta, whose opposite side is two, and\n

We call this side here a them but Tiger in\n

is seven squared. So a squared plus four is\n

the square root of 45. Since I'm worrying\n

value. Now, cosine of t is going to be adjacent\n

square root of 45 over seven. Now I go back\n

And I noticed that since I'm in the third\n

just stick a negative sign in front. This\n

same ideas as the previous solution, and ultimately\n

three properties of trig functions, the periodic\n

property. This video is about the graphs of\n

y equals cosine t and y equals sine t, where\n

the t axis, and this being the y axis. One\n

fill in this chart, using my knowledge of\n

will be easier to graph, if I convert them\n

and connect the dots to get a graph of y equals\n

pi. To continue the graph for t values less\n

more points. Or I could just use the fact\n

subtract two pi to the my angle T, I'll be\n

cosine will be exactly the same. Therefore,\n

by my y values on this graph, repeat themselves.\n

pi over six about like here, it's cosine is\n

So I'll take this dot here and repeat it over\n

plus pi over four, I get the same value of\n

this.is going to repeat. And I can continue\n

here at two pi plus, say pi over three. And\n

this. It also repeats on this side, something\n

t values will also give me the same value\n

a graph for sine and extend it by repetition.\n

sine and cosine as y equals cosine of x and\n

notice that x now refers to an angle, while\n

a different meaning of x and y, compared to\n

where x refers to the cosine value, and y\n

some properties of the graphs of sine and\n

that the graph of cosine and the graph of\n

you can think of the graph of cosine as just\n

by pi over two. So we can write cosine of\n

since adding pi over two on the inside, move\n

two. Or we can think of the graph of sine\n

by shifting the cosine graph right by pi over\n

equal to cosine of x minus pi over two, since\n

the cosine graph to the right by pi over two.

Next, let's look at domain and range. The\n

All right, that is negative infinity to infinity,\n

one. That makes sense, because sine and cosine\n

for the domain come from angles. And you can\n

as big as you want, just by wrapping a lot\n

for the range, that is the actual values of\n

on the unit circle. And those coordinates\n

than negative one. So that gives us a range.\n

from the graph. Here's cosine, that it's symmetric\n

be even. Whereas the graph of sine is symmetric\n

The absolute maximum value have these two\n

value is negative one. We can also use the\n

these two functions. The midline is the horizontal\n

points. Here, the midline is y equals zero,\n

a maximum point and the midline. You can also\n

between a minimum point and the midline, or\n

and a max point. For the cosine function and\n

periodic function is a function that repeats\n

length of the smallest repeating unit is called\n

is two pi. Notice that the period is the horizontal\n

points, or between successive troughs, or\n

cosine of x plus two pi equals cosine of x\n

to indicate that the functions repeat themselves\n

of two pi. In this video, we graphed y equals\n

that they both have a midline at y equals\n

two pi. sine u sort of functions are functions\n

like stretching and shrinking and shifting.\n

Let's start by graphing the function, y equals\n

to the function y equals sine x. So I'll graph\n

stretches this graph vertically by a factor\n

that horizontally by a factor of one half.\n

sine 2x plus one, this plus one on the outside\n

the midline amplitude and period of our original\n

sine 2x. And our further transformed y equals\n

has a midline at y equals zero, an amplitude\n

function, y equals three times sine of 2x.\n

by a factor of one half. So it changes the\n

Since the two on the inside only affects x\n

affect the midline, which is a y value, or\n

But the three on the outside does affect these\n

amplitude, since everything is stretched out\n

of one get stretch to an amplitude of three.\n

change, because multiplying a y value of zero\n

the third function, we've taken the second\n

that shifts everything up by one. Therefore,\n

we now have a midline at y equals one. The\n

three because shifting everything up by one\n

mind and the end the maximum point. Also,\n

a horizontal measure, and adding one on the\n

next, let's graph the function y equals three\n

over four. This function is very closely related\n

page, that was y equals three sine of 2x.\n

function, and maybe we can call g of x, this\n

taking f of x and plugging in x minus pi over\n

of x minus pi over four. This relationship\n

function we want to graph, we can first graph\n

page. And then we can shift its graph to the\n

you do when you subtract a number on the inside\n

three sine 2x. Recall that it's just the graph\n

three, and shrunk horizontally by a factor\n

I want, I'm going to shift this graph over\n

since I had my function written in factored\n

shift. But if I had written it instead, as\n

which is algebraically equivalent, it would\n

to shift over by pi over two. So it's best\n

the shift is, we're factoring out the coefficient\n

function, same as the one we just graphed,\n

that minus one would just bring everything\n

midline amplitude, and period for the original\n

final transformed function, y equals three\n

four minus one, our original sine function\n

one and period of two pi. For our transform\n

vertically, so it makes the amplitude three.\n

down by one. So it brings the midline, y equals\n

on the inside, shrinks everything horizontally\n

one half times two pi, which is pi. Finally,\n

function shifts to the right, by pi over four,\n

the phase shift. The function we just analyzed\n

minus one, which could also be written as\n

one. This is a function of the form y equals\n

If we have a function of this form, or the\n

know that the midline is going to be at y\n

of sine or cosine at y equals zero gets shifted\n

to be a because this A multiplied on the outside\n

to be a little more accurate, we should say\n

case a is negative. If a is negative, then\n

reflection over the x axis. We know that the\n

pi. And we know that this factor of B amounts\n

over B or I guess it could be a horizontal\n

less than one, so because we're starting with\n

by one over B, our new period is going to\n

horizontal shift. And to get that right, I\n

So instead of writing y equals a cosine bx\n

A cosine B times quantity x minus c over b\n

I write y equals a sine B times x minus quantity\n

shift as C over B. And that'll be a shift\n

a shift to the left, if C over b is negative,\n

used to, but it's because we have that minus\n

actually subtracting on the inside. So that\n

minus a negative is actually adding something,\n

as one final example, say I wanted to graph\n

minus five, that would have a midline at y\n

period of two pi divided by one half, which\n

rewrite this horizontal shift of six units\n

called the phase shift. And that's all for\n

is about graphing the trig functions, tangent,\n

intuition for the graph of y equals tangent\n

of a line at angle theta on the unit circle.\n

run. But the rise is given by sine of theta,\n

the slope is given by sine theta over cosine\n

I want to graph y equals tan of x, I can think\n

slope of the line at that angle. Notice if\n

as the angle increases towards pi over two,\n

infinity. As the angle goes from zero towards\n

negative and heading towards negative infinity\n

two, we have a vertical line. And so the slope\nis undefined.

Using this information, let's graph a rough\n

thinking of x as the angle and y as the slope,\n

pi over two and pi over two. So we said that\n

and then it heads up towards positive infinity\n

over two with an undefined value at pi over\n

infinity as the angle heads towards negative\n

at negative pi over two. You can also verify\n

two, we have the same line as for angles that\n

therefore this picture repeats and it turns\n

to pi, like sine and cosine, but just pi,\n

take a line and rotate it by 180 degrees,\n

therefore has the same value of tangents.\n

the x intercepts, all right values of x have\n

pi, two pi, etc, you can write that as pi\n

positive or negative whole number or zero.\n

of x is sine of x over cosine of x. And so\n

y is zero, which is where the numerator is\n

form pi, two pi, and so on. From the graph,\n

values like negative three pi over two, negative\n

two, these values can be written as pi over\n

this makes sense from the definition of tangents\n

the denominator is zero, and cosine x is zero,\n

two, three pi over two, and so on, the domain\n

So that's going to be everything except for\n

as x such that x is not equal to pi over two\n

or the y values go all the way from negative\n

mentioned previously, is pi. Since the smallest\n

to graph y equals secant x, I'm going to remember\n

with a graph of cosine, I can take the reciprocal\n

the reciprocal of one is one, the reciprocal\n

have a value at pi over two, negative pi over\n

pi over two. When I take the reciprocal of\n

get numbers just greater than one, but I would\n

close to zero, I'm going to get really big\n

Similarly, on the other side, over here, I\n

so their reciprocals will be negative numbers\n

of negative one is negative one. And similarly\n

negative buckets and upside down buckets as\n

Notice that secant has a period of two pi,\n

of two pi, it has a range that goes from negative\n

one to infinity. That makes sense because\n

one, and we're taking the reciprocal of those\n

the vertical asymptotes. Now the vertical\n

is at values of the form pi over two, three\n

pi over 2k, where k is an odd integer. So\n

x is not equal to pi over 2k. For K and odd\n

it doesn't have any, because you can't take\n

for your y value. We've seen the graph of\n

is the graph of y equals cotangent x. It looks\n

a decreasing function instead of an increasing\n

it's x intercepts in different places. Finally,\n

cosecant x. It's related to the graph of sine\n

in fact, if I draw the graph of sine x in\n

off. Because it's the reciprocal. I encourage\n

graphs, you can always figure out the details\n

to the graphs of cosine of x, and sine x.\n

equations. Let's start with the equation two\n

all the solutions in the interval from zero\n

for all solutions, not just those in that\n

to isolate the tricky part, which is cosine\n

negative one, and then divide both sides by\n

zero and two pi, whose cosine is negative\n

the special values on the unit circle, I can\n

that the angle between zero and two pi must\n

three, my answer needs to include both of\n

the unit circle whose cosine is negative one\n

can always take one of these angles and add\n

find all solutions, I can take these two principles\n

over three, and simply add multiples of two\n

plus two pi, or two pi over three minus two\n

on. A much more efficient way to write this\n

times k, any integer that is any positive\n

I can write four pi over three plus two pi\n

principal solution for pi over three by adding\n

This is my final solution. Next, let's look\n

usual, I'm going to start out by cleaning\n

in this case is tangent. So let me add tangent\n

x equals the square root of three. And so\n

The square root of three over three looks\n

square to three over two, which is a special\n

that my unit circle will again help me find\n

that tan x is sine x over cosine x. So I'm\n

zero and two pi with a ratio of sine over\n

three, I actually only need to look in the\n

those are the quadrants where a tangent is\n

angles whose either sine or cosine has a squared\n

see that tan pi over six, which is sine pi\n

me one half over root three over two That's\n

three, which is one over root three. If I\n

so that value works. If I try tan of pi over\n

not equal to root three over three. So pi\n

work out some the values in the third quadrant,\n

four pi over three does not. So my answer\n

over six, and seven pi over six. Now if I\n

in the interval from zero to two pi, I noticed\n

and add multiples of two pi to it, because\n

over six plus two pi k, and pi over six, sorry,\n

This is a correct answer. But it's not as\n

over six over here on the unit circle is exactly\n

both of these and adding multiples of two\n

by just taking one of them and adding multiples\n

to say that x equals pi over six, plus pi\n

capture all the same solutions. Because when\n

And when k is odd, I'll get this family. For\n

times pi is just the original seven pi over\n

has a period of pi, instead of two pi, it\n

to write the solutions in this form. In this\n

isolating sine, or tangent, or the same thing\n

unit circle, to find principal solutions.\n

zero and two pi. And then adding multiples\n

get all solutions. For tangent, we noticed\n

solution and add multiples of pi instead of\n

idea of the derivative using graphs, secant\n

So I have a function here, drawn in black,\n

here f of x equals x squared. I also have\n

This tangent line is the tangent line at the\n

a line that touches the graph of my function\n

direction as the function. Well, normally\n

But for the tangent line, we really only have\n

could approximate the slope by guessing the\n

line. But in the long run, we'll end up with\n

else. So what we're going to do instead is\n

line. A secant line is a line that goes through\n

secant line is going through my original point,\n

that's the point three, three squared or three\n

the slope of my secant line, I use the fact\n

it's the change in y over the change in x.\n

notation. F of three minus f of 1.5 is giving\n

gives me the change in x. Wilson's f of x\n

squared minus 1.5 squared over three minus\n

1.5, which ends up as 4.5. So 4.5 is the slope\n

that the slope of my secant line is an approximation\n

example, really the secant line that I've\n

approximation of the tangent line, not very\n

approximation of the slope of my tangent line?\n

as my second point, instead of using this\n

to my first point. So for example, I could\n

the point two, four, as my second point for\n

me calculate its slope, which after some arithmetic\n

I could continue to pick second points for\n

point. And I should end up with more and more\n

let me make a little chart for this. For my\n

something pretty close to 1.5, say 1.6. And\n

To take my second point to be 1.51, that's\n

To write this more generally, if I take my\n

line is going to be given by again the rise\n

minus f of 1.5 divided by x minus 1.5. Change\n

Now, there's no reason I necessarily have\n

side of my first point, I could be using stead\n

my chart, letting my second point be one,\n

of a secant line of 2.5. Here, I could get\n

1.4 and get a slope of 2.9, and so on. In\n

use a secant line through that point in our\n

in y over change in x, which is going to be\n

Actually, I can rewrite this a little bit\n

here. If I multiply the numerator and denominator\n

f of x minus f of 1.5 divided by x minus 1.5.\n

the same. So the only difference here on my\n

x as being a little bit bigger than 1.5. And\n

than 1.5. But I get the exact same expression\n

Now, this process of picking points closer\n

left, and from the right, should remind you\n

line is the limit as x goes to 1.5 of the\n

by this expression. This quantity is so important\n

derivative of f of x at x equals 1.5. So in\n

as f prime at 1.5, is the limit as x goes\n

1.5. Now based on our numerical tables, for\n

seems to be heading towards three, whether\n

left. So I'll write down the answer of three.\n

to have a really precise argument. We'd actually\n

Exactly using the formula for the function\n

do examples like that in a future video. But\n

of the tangent line is the limit of the slope\n

formula. For now, let's look at an animation\n

approach the slope of our tangent line. So\n

x squared. The red line is the tangent line\n

the blue line is a secant line that goes through\n

point with x coordinate 2.5. The points are\n

this slider here and drag my second point\n

the x coordinate, my second point gets closer\n

closer and closer to my tangent line. So the\n

of the slope of my secant lines, as my x coordinate\n

true, even if I start with my second point\n

that second point closer and closer to the\n

gets closer and closer to the slope of that\n

the slope of our tangent line, or the derivative\n

1.5 of f of x minus f of 1.5 divided by x\n

of a function y equals f of x at an x value\n

as x goes to a of f of x minus F of A over\n

at A. If this limit exists. In particular,\n

the right have to exist and be equal for the\n

There's another equivalent version of the\n

and very useful. If we're looking at the graph\n

slope of the secant line between the points\n

the letter H to be the quantity x minus a.\n

the slope of this secant line, I can write\n

a plus H. And so I can rewrite the definition\n

A equals the limit as x goes to a of f of\n

substituting in this expression for x. And\n

a is going to zero. In other words, h is going\n

as h goes to zero of f of a plus h minus f\n

we're just relabeling this point right here,\n

the slope of the tangent line is still the\n

goes to zero, this is the definition of derivative\n

when we actually calculate derivatives based\n

now, let's look at some examples to practice\n

it. So each of these following two expressions\n

some function at some value a. So for each\n

and figure out the value of A. Now remember,\n

on. They're both equivalent, but they look\n

of f at a is the limit as x goes to a of f\n

other version is the limit as h goes to zero\n

you might notice That our first expression\n

x is going to some number that's not zero.\n

here. Whereas our second expression looks\n

the age going to zero, and we've just got\n

let's look at this first one here. First,\n

like a has got to be negative one. Since x\n

makes sense, because now here on the denominator,\n

negative one. So that's our x minus a with\n

a, now we need to find an F. And we will need\n

f of a, well, let's try the simplest thing,\n

squared, then X plus five squared is our f\n

is going to be negative one plus five squared,\n

we've got our f of x here, our f of a here,\n

the definition of derivative done the first\n

we need to figure out what a is. And we need\n

the expression here is supposed to be f of\n

here, that f of x should be three to some\n

how that works. Now we need this nine to be\n

And the only way that can work is if a is\n

a plus h, that is f of two plus h

two B three to the two plus H and that actually\n

the x, it's all falling into place. So we've\n

h. And it all works where f of x being three\n

introduced the idea of derivative as a slope\n

definitions of the derivative in terms of\n

of derivatives in the next video. This video\n

with calculating derivatives using the limit\n

actually two versions of the limit definition\n

the limit as h goes to zero of f of a plus\n

you can try reworking the problems using the\n

limit as x goes to a of f of x minus F of\n

of f of x, which is one over the square root\n

Well, in other words, we want to find f prime\n

goes to zero of f of negative one plus h minus\n

of f, that's the limit of one over the square\n

one over the square root of three minus negative\n

So this is one over the square root of, let's\n

three plus one, or four minus h minus one\n

over h. And I guess I can replace the square\n

Now, unfortunately, I can't just evaluate\n

if I try that, I get one of these zero over\n

these a lot when calculating derivatives by\n

remember the context where computing slopes\n

and closer together. So our rise and our runs\n

we'll get these zero or zero indeterminate\n

that we learned before, for rewriting our\n

the limit. And I see two things going on here,\n

also fractions. So it's anybody's guess which\n

for square roots, which would be multiplying\n

the trick for fractions, which would be adding\n

I guess I'll try my fraction trick first.\n

here, is just the product of these two denominators.\n

times two. Let me rewrite my fractions with\n

I get the limit of two minus the square root\n

h times two, all over h. instead of dividing\n

see here, let's see if we can evaluate by\n

when I try to plug in, I'm still getting the\n

out of tricks, I haven't used the conjugate\n

and the bottom by the conjugate of the top.\n

two square to four minus h minus two square\n

four minus h squared, that's going to cancel\n

h squared of four minus h times two plus square\n

for now. That's good thing, I have unlimited\n

numerator, I'm going to get four minus four\n

for the ride now, oh, I see something good.\n

minus four plus H. Subtracting out those fours\n

that divide by each other, I think I finally\n

getting a zero over zero indeterminate form.\n

to zero, I'm just going to get one over two\n

a squared of four, which equals 1/16. So by\n

problem was, I think I have it, let's go back\n

of f of x, which was one over a squared of\n

set up the limit definition, did a bunch of\n

using the conjugate trick, and eventually\n

algebra doesn't get much harder than this\n

asked to find the equation of the tangent\n

two. So the slope of the tangent line is given\n

calculate the derivative first. f prime of\n

two plus h minus f of two, all over h. So\n

three times two plus h minus

two cubed minus three times two. I'm just\n

definition of my function, and then I'm plugging\n

over h. Once again, if I try to plug in zero\n

get something that all cancels out to zero\n

one of my classic zero over zero indeterminate\n

simplify things. And hope I can calculate\n

simplifying here is to multiply out two plus\n

three times two squared times h, plus three\n

getting this from the formula for multiplying\n

can also get it more slowly just by writing\n

and distributing. Now I need to subtract three\n

I needed to subtract my two cubed and then\n

if you have my terms cancel out to zero here,\n

got a minus three times two and a plus three\n

are left have H's and so I'm going to factor\n

here. And that gives me let's say, three times\n

h squared minus three over h. Now, h divided\n

as h goes to zero of 12 plus six h plus h\n

can just plug in H zero, and I get 12 minus\n

line, my derivative is nine. I'm not quite\n

the tangent line, I just know that its slope\n

equation of any line is something like y equals\n

equals 9x plus b, I just need to find the\n

I need a plug in a point. what point do I\n

talking about a tangent line here. So we've\n

x equals two, and the corresponding y value\n

or two. So my tangent line has to go through\n

this point for x and y, I get two equals nine\n

equal negative 16. So the equation of my tangent\n

found that by first calculating the derivative\n

of tangency, plugging in the x value to get\n

to finish off the equation. So in this video,\n

to compute some derivatives, using the definition\n

So fortunately, pretty soon, we'll learn some\n

without resorting to the definition. We've\n

f of x at a point, x equals A represents the\n

A f of a. But if the function f of x represents\n

a function of time, or fuel efficiency as\n

also have a practical interpretation. This\n

One of the most famous contexts for interpreting\n

So let's say I'm on a bike ride, heading straight\n

y equals f of x represents my distance from\n

f of x is my distance and miles, the distance\n

It's kind of fun to see what this graph here\n

what's going on up here, where my y values\n

where my function is constant. Please pause\n

make up a story that fits the graph. So here\n

it's actually starting to decrease. So I must\n

campus again, over here, where my f of x is\n

to take a rest or maybe I'm fixing a flat\n

Consider these two points, three, f of three,\n

slope of the secant line through those two\n

x. And y here is distance and x here is time.\n

Sounds a lot like speed, or more accurately,\n

direction, and is positive. If distance is\n

Speed is the absolute value of velocity, and\n

case, the velocity here must be negative,\n

could estimate it very roughly as about, say\n

negative eight miles per hour. But what is\n

Since we're looking at the change in distance,\n

of my secant line gives my average velocity\n

velocity, and exactly three hours or exactly\n

want to know my exact velocity at exactly\n

slope of the tangent line at x equals three.\n

sometimes called the instantaneous velocity\n

over a time interval. Let's think for a minute\n

is given by the slope of the tangent line.\n

of the tangent line is the limit of the slope\n

as x goes to three of f of x minus f of three\n

ratios represents an average velocity on the\n

is the limit of average velocities on tinier\n

one second 100th of a second, in the limit,\n

zero, we're going to get the exact velocity\n

example, the slope of the secant line represents\n

the derivative at x equals three, written\n

of the tangent line, that derivative represents\n

More generally, if f of x represents any quantity\n

line represents an average rate of change.\n

of a represents an instantaneous rate of change.\n

examples. Let's suppose that f of x represents\n

Fahrenheit as a function of time and minutes\n

interpret the first equation. f of zero is\n

the temperature is 140 degrees. What about\n

20. That's saying that the temperature goes\n

zero to 10 minutes. Now, what about this quotient\n

quotient looks a lot like the slope of a secant\n

of change. And in this context, we have the\n

two degrees per minute, as x changes from\n

of f at 15 is negative point five means that\n

decreasing at a rate of point five degrees\n

decreasing, and f prime is an instantaneous\n

Please pause the video and try this one for\n

efficiency of a Toyota Prius and mpg as a\n

that is traveling g of 45 is 52 means that\n

is 52 miles per gallon. The second statement\n

to 45 miles per hour, fuel efficiency goes\n

third statement says that the average rate\nof change

of fuel efficiency is two miles per gallon\n

35 to 40 miles per hour. So going up from\n

here. On the other hand, when you're going\n

decreasing at a rate of two miles per gallon\n

efficiency here occurs somewhere between 40\n

interpreted the slope of the secant line as\n

of the tangent line with the derivative as\n

these general principles will help you interpret\n

you might encounter throughout your life.\n

of a function as being a function in itself,\n

of its derivative, and we'll talk about where\n

that for a function f of x and a number A,\n

given by this formula. But what if we let\n

different values of a, we can think of the\n

I'm going to rewrite this definition of derivative\n

a little more like standard function notation.\n

as h goes to zero of f of x plus h minus f\n

different from what we've been doing before,\n

let's do one more example of computing the\n

a general number x instead of a specific value,\n

equals one over x. And first, let's just write\n

So f prime of x is given by this formula.\n

x, I can rewrite this as one over x plus h\n

this is a zero over zero indeterminate form.\n

get one over x minus one over x on the numerator,\n

gives me zero on the denominator two. So I'll\n

to get an A form that I can evaluate it. Let's\n

here. The common denominator I need to use\n

by x over x and the next fraction by x plus\n

continuing, I get x minus x plus h over x\n

whole thing by h here, multiply by one over\n

over x plus h times x times h. Now I can subtract\n

divide my minus h by my H, to get just a minus\n

the limit of negative one over x plus h times\n

I can plug in H equals zero and get something\n

of negative one over x plus zero times x or\n

In this example, we're given the graph of\n

height of an alien spaceship above the Earth's\n

The rate of change means the derivative of\n

to work with. So we'll just have to estimate\n

by thinking about slopes of tangent lines.\n

I can graph my derivative. And I'll consider\n

For x values between zero and two, my original\n

slope negative one, since the rise is negative\n

the straight line segment, the tangent line\n

one, and therefore, the derivative will be\n

two, I'm going to ignore the time being what\n

two, and just look at the interval of X values\n

flat. So tangent lines at any of these points\n

of zero. When x is between two and three,\n

when x is exactly three. And just think about\n

five, where g of x is flat again, so it's\n

I'll draw again, a derivative is zero when\n

a little more interesting. As x increases\n

function. The slope of tangent lines here\n

of three, and decreasing to a slope of zero,\n

As x increases from seven, the tangent lines\n

negative slope of about negative one here,\n

when x is just shy of 10. My estimates of\n

tangent lines are just rough estimates based\n

x increases from 10, the tangent line slope\n

so my derivative is going to be positive and\n

derivative. Now let's see what happens at\n

figure out the derivative at x equals two,\n

as the limit of the slopes of the secant lines.\n

the left, I'll just get this line that lines\n

one. But if I compute the slope of a secant\n

a slope of zero. So the limit from the left\n

of my secret lines will be different. And\n

does not exist. And so I'll just draw this\n

look at the derivative when x equals three.\n

limit as h goes to zero, of g of three plus\n

is bigger than zero, then g of three plus\n

plus H is to the right of three. On the other\n

plus H is two, because three plus H is actually\n

is equal to one half, based on the filled\n

as h goes to zero from the positive side,\n

over age, which is just the limit of zeros,\n

the limit from the left, we get the limit\n

limit of three halves over h. And as h goes\n

So once again, the left limit and write limit\n

of the secant lines does not exist, and there's\n

draw an open circle there to

add x equals five, again, we have a corner.\n

that the derivative does not exist. And finally,\n

from the right not the left. And so by that\n

derivative at that left endpoint either. So\n

of the height of our alien spaceship, as it\n

Earthlings and then makes us escape up to\n

that the domain of the original function g\n

of g prime is somewhat smaller, and just goes\n

then from three to five, and finally, from\n

the function originally existed. We saw in\n

doesn't necessarily exist at all the x values\n

Please pause the video for a moment and try\n

you can, that a derivative can fail to exist

at an x value x equals a. Well, one kind of\n

have a derivative at x equals A is if f of\n

example, if it has a hole, like in this picture,\n

spacecraft, that a derivative can fail to\n

we tried to evaluate the derivative in that\n

of the secant lines, the limit from the left,\n

A famous example of a function that turns\n

the absolute value function, f prime of x\n

one, when x is less than zero, and it's positive\n

of zero itself does not exist. A function\n

at the cusp. In the alien spaceship example,\n

at a discontinuity. But there's another way\n

when f has no cost per corner discontinuity.\n

to the 1/3 graphed here, what's going on at\n

line is a vertical with a slope that's infinite\n

the secant lines will fail to exist because\n

at x equals a, if the derivative exists at\n

if f is differentiable at every point in that\n

previous slides are examples of places where\n

examples are important. But I'm going to focus\n

general, if f of x is not continuous at x\n

x equals a. This is what we saw in the example\n

way of saying the same thing is that f is\n

previous slides are examples of places where\n

examples are important. But I'm going to focus\n

general, if f of x is not continuous at x\n

x equals a. This is what we saw in the example\n

way of saying the same thing is that f is\n

However, if all we know is that f is continuous\n

about whether or not is differentiable there,\n

a. Remember the square root example, the square\n

but it's not differentiable there because\n

graph of a function to the graph of its derivative.\n

we also looked at several ways that a derivative\n

if a function is differentiable, it has to\n

differentiable functions are continuous. What\n

is differentiable at a number x equals a,\n

call the function f of x. And I'm going to\n

f of x to be differentiable at x equals a.\n

f of x minus F of A over x minus a exists\n

However, if all we know is that f is continuous\n

about whether or not is differentiable there,\n

a. Remember the square root example, the square\n

but it's not differentiable there because\n

graph of a function to the graph of its derivative.\n

we also looked at several ways that a derivative\n

if a function is differentiable, it has to\n

differentiable functions are continuous. What\n

is differentiable at a number x equals a,\n

call the function f of x. And I'm going to\n

f of x to be differentiable at x equals a.\n

f of x minus F of A over x minus a exists\n

of a. Now I'm going to multiply both sides\nof this equation

of a. Now I'm going to multiply both sides\nof this equation

by the limit as x goes to a of x minus a.\n

on to make sure this is legit. Does this limit\n

as x goes to a of x exists, that's just a,\n

that's a also. So the limit of the difference\n

the difference is the difference of the limits,\n

So I've actually just multiplied both sides\n

by the limit as x goes to a of x minus a.\n

on to make sure this is legit. Does this limit\n

as x goes to a of x exists, that's just a,\n

that's a also. So the limit of the difference\n

the difference is the difference of the limits,\n

So I've actually just multiplied both sides\n

This is actually a surprisingly useful thing\n

of two limits here, both of which exist. So\n

this as the limit as x goes to a

This is actually a surprisingly useful thing\n

of two limits here, both of which exist. So\n

this as the limit as x goes to a

of x minus a times f of x minus F of A over\n

of x minus a, which is fine to do when x is\n

the limit of f of x minus f of a is equal\n

limit was just zero. So my limit on the left\nis equal to zero.

of x minus a times f of x minus F of A over\n

of x minus a, which is fine to do when x is\n

the limit of f of x minus f of a is equal\n

limit was just zero. So my limit on the left\nis equal to zero.

And now I'm so close, I'd like to apply five\n

it up into a difference of limits, but I can't\n

the limit, as x goes to a of f of x exists,\n

prove as far as continuity. So instead, I\n

that I do know exists. And that's the limit

And now I'm so close, I'd like to apply five\n

it up into a difference of limits, but I can't\n

the limit, as x goes to a of f of x exists,\n

prove as far as continuity. So instead, I\n

that I do know exists. And that's the limit

Now, I do know that both of these two limits\n

limit rule about sums to rewrite this limit.

Now, I do know that both of these two limits\n

limit rule about sums to rewrite this limit.

Now, on the left side, I can cancel out my\n

and I get that the limit as x goes to a of\n

rule for psalms that I applied

Now, on the left side, I can cancel out my\n

and I get that the limit as x goes to a of\n

rule for psalms that I applied

That limit their past to equal the limit as\n

some number, doesn't matter what X is doing\n

is. So this limit on the right is just f of\n

means for a function to be continuous at x\n

x equals f of a. So f is continuous at x equals\n

we prove that if f is differentiable at x\n

a. This statement is equivalent to another\n

which says that if f is not continuous, at\n

x equals a. In this video, we'll learn a few\n

the power role, and the derivatives of sums,\n

rules will give us shortcuts for finding derivatives\n

old limit definition of derivative. In this\n

and some examples, there won't be any proofs\n

these proofs will be in a separate later video.\n

if we have a constant C, and we will have\n

that, it's just going to be a straight horizontal\nline.

That limit their past to equal the limit as\n

some number, doesn't matter what X is doing\n

is. So this limit on the right is just f of\n

means for a function to be continuous at x\n

x equals f of a. So f is continuous at x equals\n

we prove that if f is differentiable at x\n

a. This statement is equivalent to another\n

which says that if f is not continuous, at\n

x equals a. In this video, we'll learn a few\n

the power role, and the derivatives of sums,\n

rules will give us shortcuts for finding derivatives\n

old limit definition of derivative. In this\n

and some examples, there won't be any proofs\n

these proofs will be in a separate later video.\n

if we have a constant C, and we will have\n

that, it's just going to be a straight horizontal\nline.

df dx, ought to be zero, because the slope\n

df dx, ought to be zero, because the slope\n

Another simple example, is the derivative\n

if I draw the graph, that's just going to\n

the tangent line for the straight line will\n

prime of x has to be always equal to one.\n

cases of the power rule, which is one of the\n

Another simple example, is the derivative\n

if I draw the graph, that's just going to\n

the tangent line for the straight line will\n

prime of x has to be always equal to one.\n

cases of the power rule, which is one of the\n

So the power rule says that if you have the\n

any real number, then you can find the derivative\n

down and multiplying it in the front and then\n

if you want to calculate the derivative of\n

going to be 15 times x to the 15 minus one,\n

cube root of x might not immediately look\n

rule. But if we rewrite it, by putting the\n

1/3, now we can apply the power rule, we bring\n

So the power rule says that if you have the\n

any real number, then you can find the derivative\n

down and multiplying it in the front and then\n

if you want to calculate the derivative of\n

going to be 15 times x to the 15 minus one,\n

cube root of x might not immediately look\n

rule. But if we rewrite it, by putting the\n

1/3, now we can apply the power rule, we bring\n

exponent of 1/3 by one, or 1/3 minus one is\n

here using the power rule, we could rewrite\n

one over 3x to the two thirds, either answers\ngood.

exponent of 1/3 by one, or 1/3 minus one is\n

here using the power rule, we could rewrite\n

one over 3x to the two thirds, either answers\ngood.

In the third example, g of x is one over x\n

rewriting before we can apply the power rule.\n

minus 3.7. Using exponent rules, now I can\n

multiplying in the front, and now I have to\n

one that gives me x to the negative 4.7. Again,\n

over x to the 4.7. It's important to notice\n

any example where the power rule applies,\n

is just a constant, just a real number. The\n

just a constant real number, and f is a differentiable\n

of x is just c times the derivative of f of\n

we can just pull a constant outside of the\n

example. If we want to take the derivative\n

times the derivative of x cubed. And now using\n

and get 15x squared. f and g are differentiable\n

g of x is the derivative of f plus the derivative\n

g are differentiable functions, then the derivative\n

the derivatives. Now let's use all these rules\n

polynomial. To find the y dx, we can use the\n

of each term separately. Now using the constant\n

out the seven, bring down the three getting\n

five times two times x to the one, four times\n

the derivative of a constant two is just zero.\n

plus four, and notice that the derivative\n

polynomial of one less degree. In this video,\n

of various functions, especially polynomials.\n

rules come from, how they're derived from\n

for another video coming soon on proofs. This\n

like sine and cosine. But I want to start\n

In the third example, g of x is one over x\n

rewriting before we can apply the power rule.\n

minus 3.7. Using exponent rules, now I can\n

multiplying in the front, and now I have to\n

one that gives me x to the negative 4.7. Again,\n

over x to the 4.7. It's important to notice\n

any example where the power rule applies,\n

is just a constant, just a real number. The\n

just a constant real number, and f is a differentiable\n

of x is just c times the derivative of f of\n

we can just pull a constant outside of the\n

example. If we want to take the derivative\n

times the derivative of x cubed. And now using\n

and get 15x squared. f and g are differentiable\n

g of x is the derivative of f plus the derivative\n

g are differentiable functions, then the derivative\n

the derivatives. Now let's use all these rules\n

polynomial. To find the y dx, we can use the\n

of each term separately. Now using the constant\n

out the seven, bring down the three getting\n

five times two times x to the one, four times\n

the derivative of a constant two is just zero.\n

plus four, and notice that the derivative\n

polynomial of one less degree. In this video,\n

of various functions, especially polynomials.\n

rules come from, how they're derived from\n

for another video coming soon on proofs. This\n

like sine and cosine. But I want to start\n

quadratic functions. If I want to find the\n

it x squared minus 6x minus seven equals zero,\n

equals zero, set the factors equal to 0x minus\n

And that gives me the solutions, x equals\n

look at this more complicated equation. I'm\n

out the right hand side. Next, our combined\n

me x squared minus 6x on both sides, well,\n

6x. That's true no matter what I plug in for\n

this equation, we can say that the solution\n

is called an identity, because it holds for\n

on the other hand is not an identity, because\n

all values. Please pause the video for a moment\n

equations or identities that is, which of\n

variable. To start out, you might want to\n

variable and see if the equation holds. The\n

hold for some values of x. For example, if\n

is zero. Two times sine of zero is also zero.\n

x is say pi over two, then sine of two times\n

of pi, which is zero, but two times sine of\n

and zero is not equal to two. So the equation\n

quadratic functions. If I want to find the\n

it x squared minus 6x minus seven equals zero,\n

equals zero, set the factors equal to 0x minus\n

And that gives me the solutions, x equals\n

look at this more complicated equation. I'm\n

out the right hand side. Next, our combined\n

me x squared minus 6x on both sides, well,\n

6x. That's true no matter what I plug in for\n

this equation, we can say that the solution\n

is called an identity, because it holds for\n

on the other hand is not an identity, because\n

all values. Please pause the video for a moment\n

equations or identities that is, which of\n

variable. To start out, you might want to\n

variable and see if the equation holds. The\n

hold for some values of x. For example, if\n

is zero. Two times sine of zero is also zero.\n

x is say pi over two, then sine of two times\n

of pi, which is zero, but two times sine of\n

and zero is not equal to two. So the equation\n

Since it doesn't hold for all values of the\n

equation is an identity. You can build some\n

For example, cosine of zero plus pi, which\n

of cosine of zero. You can also check for\n

is the same thing as negative cosine of pi\n

that's just evidence, it's not a proof that\n

lucky with the values we picked, we can build\n

at graphs, I'm going to put theta on the x\n

Since it doesn't hold for all values of the\n

equation is an identity. You can build some\n

For example, cosine of zero plus pi, which\n

of cosine of zero. You can also check for\n

is the same thing as negative cosine of pi\n

that's just evidence, it's not a proof that\n

lucky with the values we picked, we can build\n

at graphs, I'm going to put theta on the x\n

that's just like the graph of cosine shifted\n

if I graph y equals negative cosine theta,\n

across the x axis, which gives us the exact\n

strong evidence that this equation is an identity,\n

strongest evidence of all would be an algebraic\n

once we have a formula for the cosine of a\nsum of two angles.

that's just like the graph of cosine shifted\n

if I graph y equals negative cosine theta,\n

across the x axis, which gives us the exact\n

strong evidence that this equation is an identity,\n

strongest evidence of all would be an algebraic\n

once we have a formula for the cosine of a\nsum of two angles.

In the meantime, let's look at equation C.\n

we could build evidence for it again by plugging\n

and the right side separately, and checking\n

this example, I'm going to go ahead and do\n

In the meantime, let's look at equation C.\n

we could build evidence for it again by plugging\n

and the right side separately, and checking\n

this example, I'm going to go ahead and do\n

In particular, I'm going to start with the\n

and rewrite things until I get to the right\n

In particular, I'm going to start with the\n

and rewrite things until I get to the right\n

is secant and tangent in terms of their constituent\n

x is one over cosine x, and tangent of x is\n

as one over cosine x minus sine x times sine\n

and write this as one over cosine x minus\n

that I have two fractions with the same denominator.\n

squared x over cosine x. Next, I'm going to\n

x using the Pythagorean identity that says\n

equals one, and therefore, one minus sine\n

by subtracting sine squared x from both sides.\n

squared x with cosine squared x. And canceling\n

that's the same thing as cosine of x, which\n

get to. So a combination of a bunch of algebra,\n

prove that this equation is true for all values\nof x

is secant and tangent in terms of their constituent\n

x is one over cosine x, and tangent of x is\n

as one over cosine x minus sine x times sine\n

and write this as one over cosine x minus\n

that I have two fractions with the same denominator.\n

squared x over cosine x. Next, I'm going to\n

x using the Pythagorean identity that says\n

equals one, and therefore, one minus sine\n

by subtracting sine squared x from both sides.\n

squared x with cosine squared x. And canceling\n

that's the same thing as cosine of x, which\n

get to. So a combination of a bunch of algebra,\n

prove that this equation is true for all values\nof x

it's an identity. The best way to prove that\n

it's an identity. The best way to prove that\n

is to use algebra and to use other identities,\n

side of the equation till it looks like the\n

equation is not an identity is to plug in\n

the equation not true. Now if you're just\n

or not, and not worried about proving it,\n

the left and right sides to see if those graphs\n

equation that holds for all values of the\n

identities called the Pythagorean identities.\n

theta plus sine squared theta equals one.\n

one equals secant squared theta. And the third\n

equals cosecant squared theta. Let's start\n

sine squared theta equals one. I'll do this\n

inside it by the definition of sine and cosine,\n

theta and sine theta, the high partners, my\n

my unit circle. Now the length of the base\n

coordinate of this point. So that's equal\n

is the same thing as the y coordinate of this\n

theorem for right triangles, says this side\n

is equal to the hypothenar squared. So by\n

theta squared plus sine theta squared equals\n

squared theta plus sine squared theta equals\n

squared theta is just a shorthand notation\n

proof of the first Pythagorean identity, at\n

in the first quadrant. In the case, when the\n

use symmetry to argue the same identity holds.\n

the next Pythagorean identity, tan squared\n

let's use the first without your an identity,\n

sine squared theta equals one, I'm going to\n

squared theta. Now I'm going to rewrite the\n

cosine squared theta over cosine squared theta\n

theta. Now cosine squared theta over cosine\n

the next fraction as sine of theta over cosine\n

a fraction, I can just square the numerator\n

theta is shorthand for sine of theta squares.\n

the other side of the equal sign, I can rewrite\n

Again, that's because when I square the fraction,\n

by the cosine theta squared, which is this.\n

is the same thing as tangent theta. And one\n

theta. Using the shorthand notation, that\n

data, which, after rearranging is exactly\n

proof of the third for that green identity\n

is to use algebra and to use other identities,\n

side of the equation till it looks like the\n

equation is not an identity is to plug in\n

the equation not true. Now if you're just\n

or not, and not worried about proving it,\n

the left and right sides to see if those graphs\n

equation that holds for all values of the\n

identities called the Pythagorean identities.\n

theta plus sine squared theta equals one.\n

one equals secant squared theta. And the third\n

equals cosecant squared theta. Let's start\n

sine squared theta equals one. I'll do this\n

inside it by the definition of sine and cosine,\n

theta and sine theta, the high partners, my\n

my unit circle. Now the length of the base\n

coordinate of this point. So that's equal\n

is the same thing as the y coordinate of this\n

theorem for right triangles, says this side\n

is equal to the hypothenar squared. So by\n

theta squared plus sine theta squared equals\n

squared theta plus sine squared theta equals\n

squared theta is just a shorthand notation\n

proof of the first Pythagorean identity, at\n

in the first quadrant. In the case, when the\n

use symmetry to argue the same identity holds.\n

the next Pythagorean identity, tan squared\n

let's use the first without your an identity,\n

sine squared theta equals one, I'm going to\n

squared theta. Now I'm going to rewrite the\n

cosine squared theta over cosine squared theta\n

theta. Now cosine squared theta over cosine\n

the next fraction as sine of theta over cosine\n

a fraction, I can just square the numerator\n

theta is shorthand for sine of theta squares.\n

the other side of the equal sign, I can rewrite\n

Again, that's because when I square the fraction,\n

by the cosine theta squared, which is this.\n

is the same thing as tangent theta. And one\n

theta. Using the shorthand notation, that\n

data, which, after rearranging is exactly\n

proof of the third for that green identity\n

cosine squared theta plus sine squared theta\n

sides by sine squared theta. I'll break up\n

my fractions as cosine theta over sine theta\n

squared. cosine over sine can be written As\n

as cosecant. That gives me the identity that\n

identities. The first one, we proved using\n

The second and third identities, we proved\n

algebra. The sum and difference formulas are\n

two angles, the cosine of a sum of two angles,\n

the cosine of a difference of two angles.\n

about this question. Is it true that the sine\n

the sine of B? No, it's not true. And we can\n

pi over two and B equals pi, than the sine\n

as a sine of three pi over two, which is negative\n

the sine of pi is equal to one plus zero,\n

one. So this equation does not hold for all\n

of a and b for which it does hold. For example,\n

true in general, instead, we need more complicated\n

sum of two angles A plus B is given by sine\n

B. The cosine of A plus B is given by cosine\n

remember these with a song, sine cosine cosine\n

feel free to back up the video and sing along\n

cosine squared theta plus sine squared theta\n

sides by sine squared theta. I'll break up\n

my fractions as cosine theta over sine theta\n

squared. cosine over sine can be written As\n

as cosecant. That gives me the identity that\n

identities. The first one, we proved using\n

The second and third identities, we proved\n

algebra. The sum and difference formulas are\n

two angles, the cosine of a sum of two angles,\n

the cosine of a difference of two angles.\n

about this question. Is it true that the sine\n

the sine of B? No, it's not true. And we can\n

pi over two and B equals pi, than the sine\n

as a sine of three pi over two, which is negative\n

the sine of pi is equal to one plus zero,\n

one. So this equation does not hold for all\n

of a and b for which it does hold. For example,\n

true in general, instead, we need more complicated\n

sum of two angles A plus B is given by sine\n

B. The cosine of A plus B is given by cosine\n

remember these with a song, sine cosine cosine\n

feel free to back up the video and sing along\n

sum of angles and the cosine of a sum of angles.\n

sum of angles and the cosine of a sum of angles.\n

and cosine of a difference of two angles.\n

and cosine of a difference of two angles.\n

A minus B as sine of A plus negative B. And\n

out to sine cosine plus cosine, sine. And\n

I know that cosine of negative B is cosine\n

b is negative sine of B. So I can rewrite\n

of A sine of B. Notice that this new formula\n

for the sum is just that plus sign turned\n

for cosine of A minus B, that's cosine of\n

b minus sine of A sine of negative B. Again,\n

cosine A cosine B plus sine A sine B. Once\n

exactly like the for the song, just that minus\n

a plus sign. Now let's use the angle sum formula\n

degrees. Now, 105 degrees is not a special\n

it as the sum of two special angles. I can\n

the sine of 105 degrees is the sine of 60\n

this is sine, cosine, cosine, sine. And I\n

sine of 60 degrees is root three over two\n

of 60 degrees is one half, and sine of 45\n

to root six plus root two over four. For our\n

W, given the values of cosine v and cosine\n

the first quadrant. Remember, to compute the\n

the two cosines. That wouldn't even make sense\n

point seven would give something bigger than\n

than one. Instead, we have to use the angle\n

of v plus w equals cosine, cosine, minus sine,\n

v and the cosine of W, so I could just plug\n

of v and the sine of W from the given information.\n

So here, I'm going to draw a right triangle\nwith angle V

A minus B as sine of A plus negative B. And\n

out to sine cosine plus cosine, sine. And\n

I know that cosine of negative B is cosine\n

b is negative sine of B. So I can rewrite\n

of A sine of B. Notice that this new formula\n

for the sum is just that plus sign turned\n

for cosine of A minus B, that's cosine of\n

b minus sine of A sine of negative B. Again,\n

cosine A cosine B plus sine A sine B. Once\n

exactly like the for the song, just that minus\n

use the angle sum formula to find the exact\n

degrees is not a special angle on the unit\n

special angles. I can write it as 60 degrees\n

degrees is the sine of 60 plus 45. And now\n

cosine, sine. And I for my Unit Circle, I\n

root three over two cosine of 45 degrees root\n

half, and sine of 45 degrees is root two over\n

two over four. For our last example, let's\n

of cosine v and cosine W, and the fact that\n

Remember, to compute the cosine of a sum,\n

That wouldn't even make sense in this case,\n

would give something bigger than one and the\n

Instead, we have to use the angle sum formula\n

w equals cosine, cosine, minus sine, sine.\n

the cosine of W, so I could just plug those\n

and the sine of W from the given information.\n

So here, I'm going to draw a right triangle\nwith angle V

and another right triangle with angle W. Since\n

I can think of that as nine over 10.

and another right triangle with angle W. Since\n

I can think of that as nine over 10.

And I can think of that as adjacent over hypotenuse\n

triangles adjacent side with the number nine\n

I know that the cosine of W is point seven,\n

this adjacent side, and a 10 on this iPod\n

compute the length of the unlabeled side.\n

of 10 squared minus nine squared, that's going\n

the square root of 10 squared minus seven\n

I can now find the sign of V as the opposite\n

root of 19 over 10. And the sign of W will\n

we're assuming v and w are in the first quadrant,\n

so we don't need to Jimmy around with positive\n

And I can think of that as adjacent over hypotenuse\n

triangles adjacent side with the number nine\n

I know that the cosine of W is point seven,\n

this adjacent side, and a 10 on this iPod\n

compute the length of the unlabeled side.\n

of 10 squared minus nine squared, that's going\n

the square root of 10 squared minus seven\n

I can now find the sign of V as the opposite\n

root of 19 over 10. And the sign of W will\n

we're assuming v and w are in the first quadrant,\n

so we don't need to Jimmy around with positive\n

Now we're ready to plug into our formula.\n

to point nine times point seven minus the\n

Now we're ready to plug into our formula.\n

to point nine times point seven minus the\n

using a calculator, this works out to a decimal\n

angle sum and difference formulas and use\n

for why the sum formulas hold, please watch\n

for sine of two theta and cosine of two theta.\n

if you think this equation sine of two theta\n

that true means always true for all values\n

always false. This equation is false, because\n

way to see this is graphically, if I graph\n

graph of sine theta, squished in horizontally\n

if I graph y equals two sine beta, that's\n

by a factor of two. These two graphs are not\n

formula for sine of two theta. And that formula\n

theta. It's not hard to see why that formula\n

that sine of A plus B is equal to sine A cosine\n

theta, which is sine of theta plus theta is\n

cosine theta sine theta. simply plugging in\n

some formula, will sine theta cosine theta\n

So I can rewrite this as twice sine theta\n

There's also a formula for cosine of two theta.\n

sine squared theta. Again, we can use the\n

from. cosine of A plus B is equal to cosine\n

if we want cosine of two theta, that's just\n

theta, cosine theta, minus sine theta, sine\n

can be rewritten as cosine squared theta minus\n

above. Now there are a couple other formulas\n

One of them is one minus two sine squared\n

theta is two cosine squared theta minus one,\n

the original one using the Pythagorean identity.\n

squared theta is one. So cosine squared theta\n

that into my original formula, which I've\n

cosine squared, I'm gonna write one minus\n

minus sine squared theta. So that's the same\n

which is exactly what I'm looking for. Similarly,\n

sine squared theta as one minus cosine squared\n

copy it below. But this time, I'm going to\n

gives me cosine of two theta is cosine squared\n

squared theta. That simplifies to two cosine\n

the negative sign and combining like terms.

using a calculator, this works out to a decimal\n

angle sum and difference formulas and use\n

for why the sum formulas hold, please watch\n

for sine of two theta and cosine of two theta.\n

if you think this equation sine of two theta\n

that true means always true for all values\n

always false. This equation is false, because\n

way to see this is graphically, if I graph\n

graph of sine theta, squished in horizontally\n

if I graph y equals two sine beta, that's\n

by a factor of two. These two graphs are not\n

formula for sine of two theta. And that formula\n

theta. It's not hard to see why that formula\n

that sine of A plus B is equal to sine A cosine\n

theta, which is sine of theta plus theta is\n

cosine theta sine theta. simply plugging in\n

some formula, will sine theta cosine theta\n

So I can rewrite this as twice sine theta\n

There's also a formula for cosine of two theta.\n

sine squared theta. Again, we can use the\n

from. cosine of A plus B is equal to cosine\n

if we want cosine of two theta, that's just\n

theta, cosine theta, minus sine theta, sine\n

can be rewritten as cosine squared theta minus\n

above. Now there are a couple other formulas\n

One of them is one minus two sine squared\n

theta is two cosine squared theta minus one,\n

the original one using the Pythagorean identity.\n

squared theta is one. So cosine squared theta\n

that into my original formula, which I've\n

cosine squared, I'm gonna write one minus\n

minus sine squared theta. So that's the same\n

which is exactly what I'm looking for. Similarly,\n

sine squared theta as one minus cosine squared\n

copy it below. But this time, I'm going to\n

gives me cosine of two theta is cosine squared\n

squared theta. That simplifies to two cosine\n

the negative sign and combining like terms.

So I have one double angle formula for sine\n

the double angle formula for cosine of two\ntheta.

So I have one double angle formula for sine\n

the double angle formula for cosine of two\ntheta.

Now let's use these formulas in some examples.\n

know that cosine theta is negative one over\n

three, we have a choice of three formulas\n

the second one, because it only involves cosine\n

know my value for cosine theta. Of course,\n

Now let's use these formulas in some examples.\n

know that cosine theta is negative one over\n

three, we have a choice of three formulas\n

the second one, because it only involves cosine\n

know my value for cosine theta. Of course,\n

but then I'd have to work out the value of\n

two theta is twice negative one over root\n

tenths minus one or negative eight tenths,\n

the equation two cosine x plus sine of 2x\n

is that one of the trig functions has the\n

has the argument of 2x. So I want to use my\n

I'll copy down the two cosine x, and now sine\n

this point, I see a way to factor my equation,\n

of these two terms. That gives me one plus\n

That means that either two cosine x is equal\n

that simplifies to cosine x equals zero, or\n

I see that cosine of x is zero at pi over\n

x is negative one at three pi over two, there's\n

is going to be pi over two plus multiples\n

of two pi. This video proved the double angle\n

cosine theta. and cosine of two theta is cosine\n

also proved to alternate versions of the equation\n

higher order derivatives and notation. We've\n

of the function f of x, but f prime of x is\n

derivative, that would be f prime prime of\n

prime of x. This is called the second derivative\n

we can also talk about the third derivative,\n

be written f to the three of x if you get\n

can talk about the nth derivative f, parentheses\n

to show that it's the nth derivative. The\n

but then I'd have to work out the value of\n

two theta is twice negative one over root\n

tenths minus one or negative eight tenths,\n

the equation two cosine x plus sine of 2x\n

is that one of the trig functions has the\n

has the argument of 2x. So I want to use my\n

I'll copy down the two cosine x, and now sine\n

this point, I see a way to factor my equation,\n

of these two terms. That gives me one plus\n

That means that either two cosine x is equal\n

that simplifies to cosine x equals zero, or\n

I see that cosine of x is zero at pi over\n

x is negative one at three pi over two, there's\n

is going to be pi over two plus multiples\n

of two pi. This video proved the double angle\n

cosine theta. and cosine of two theta is cosine\n

also proved to alternate versions of the equation\n

higher order derivatives and notation. We've\n

of the function f of x, but f prime of x is\n

derivative, that would be f prime prime of\n

prime of x. This is called the second derivative\n

we can also talk about the third derivative,\n

be written f to the three of x if you get\n

can talk about the nth derivative f, parentheses\n

to show that it's the nth derivative. The\n

There are many alternative notations for derivatives\n

history in the 1600s. There are a few different\n

we write a function of something like f of\n

to the output of a function. When we're looking\n

the notation f prime of x. But you might also\n

notation is df dx, is known as lug nuts denotation.\n

like dy dx of f of x. And you might see dy\n

Sometimes you'll also see a capital D used\n

at the second derivative, we've seen that\n

is a similar notation, or we might write dy\ndx of df

There are many alternative notations for derivatives\n

history in the 1600s. There are a few different\n

we write a function of something like f of\n

to the output of a function. When we're looking\n

the notation f prime of x. But you might also\n

notation is df dx, is known as lug nuts denotation.\n

like dy dx of f of x. And you might see dy\n

Sometimes you'll also see a capital D used\n

at the second derivative, we've seen that\n

is a similar notation, or we might write dy\ndx of df

And the shorthand for that is d squared f\n

y dx squared using y in the place of F for\nthe function.

And the shorthand for that is d squared f\n

y dx squared using y in the place of F for\nthe function.

There's similar notations for third derivative,\n

would be F to the n of x, or y to the n, d\n

y dx to the N. When using live minutes notation,\n

our derivative at a particular value of x,\n

three, or at x equals a, using a vertical\n

to become familiar with all of these alternative\nnotations.

There's similar notations for third derivative,\n

would be F to the n of x, or y to the n, d\n

y dx to the N. When using live minutes notation,\n

our derivative at a particular value of x,\n

three, or at x equals a, using a vertical\n

to become familiar with all of these alternative\nnotations.

That's all for this video on higher order\n

the derivative of e to the x, one of my favorite\n

a great derivative. As you may recall, he\n

is something like 2.718 looks like it's repeating.\n

That's all for this video on higher order\n

the derivative of e to the x, one of my favorite\n

a great derivative. As you may recall, he\n

is something like 2.718 looks like it's repeating.\n

Its value notice is somewhere in between two\n

Its value notice is somewhere in between two\n

It says exponential function increase thing\n

the x, not only is the graph of e to the x\n

rapidly. So for negative values of x, the\n

to zero. Over here, when x equals zero, that\n

of one, we'll see that it is in fact exactly\n

lines get steeper and steeper. I'm going to\n

about E. First, if you take the limit, as\n

raised to the nth power, that limit exists\n

like that when you took precalculus, and looked\n

and smaller time periods. But even if you\n

fact worth memorizing, you'll see it again\n

is that the limit as h goes to zero of e to\n

expression here on the left may remind you\n

as the limit as h goes to zero of e to the\n

the zero is one over h, that's equal to one.\n

is just the derivative of e to the x at x\n

of derivative. So this fact, is really saying\n

zero, that derivative is equal to one. So\nfor the third fat

It says exponential function increase thing\n

the x, not only is the graph of e to the x\n

rapidly. So for negative values of x, the\n

to zero. Over here, when x equals zero, that\n

of one, we'll see that it is in fact exactly\n

lines get steeper and steeper. I'm going to\n

about E. First, if you take the limit, as\n

raised to the nth power, that limit exists\n

like that when you took precalculus, and looked\n

and smaller time periods. But even if you\n

fact worth memorizing, you'll see it again\n

is that the limit as h goes to zero of e to\n

expression here on the left may remind you\n

as the limit as h goes to zero of e to the\n

the zero is one over h, that's equal to one.\n

is just the derivative of e to the x at x\n

of derivative. So this fact, is really saying\n

zero, that derivative is equal to one. So\nfor the third fat

it talks about the derivative of e to the\n

the derivative of the function, either the\n

is its own derivative. So this is a generalized\n

fact is saying that the derivative at x equals\n

as e to the zero. So it saying the drought\n

to zero. And in general, the derivative of\n

fact, one is frequently taken as the definition\n

as a definition of he, since he is the unique\n

number, you can plug in here and get this\n

fact one implies fact two and vice versa.\n

to prove that fact two implies fact three\n

pretty straightforward from the definition\n

So let's start out assuming fact to and try\n

direct by the definition of derivative. The\n

goes to zero of e to the x plus h minus e\n

X, from both terms on the numerator, I get\n

one over h. Notice that e to the x times e\n

rules. Now, either the X has nothing to do\n

H is concerned. And I can pull it all the\n

limit. Now by factor two, which I'm assuming\n

it talks about the derivative of e to the\n

the derivative of the function, either the\n

is its own derivative. So this is a generalized\n

fact is saying that the derivative at x equals\n

as e to the zero. So it saying the drought\n

to zero. And in general, the derivative of\n

fact, one is frequently taken as the definition\n

as a definition of he, since he is the unique\n

number, you can plug in here and get this\n

fact one implies fact two and vice versa.\n

to prove that fact two implies fact three\n

pretty straightforward from the definition\n

So let's start out assuming fact to and try\n

direct by the definition of derivative. The\n

goes to zero of e to the x plus h minus e\n

X, from both terms on the numerator, I get\n

one over h. Notice that e to the x times e\n

rules. Now, either the X has nothing to do\n

H is concerned. And I can pull it all the\n

limit. Now by factor two, which I'm assuming\n

is e to the x, just like I wanted to show.\n

to compute the derivative of a function that\n

is e to the x, just like I wanted to show.\n

to compute the derivative of a function that\n

combined in lots of different ways. You'll\n

of e to the x that we just talked about, but\n

combined in lots of different ways. You'll\n

of e to the x that we just talked about, but\n

So please pause the video and try to compute\n

to what's a variable and what's the constant.

So please pause the video and try to compute\n

to what's a variable and what's the constant.

Okay, so we're taking the derivative here\n

I'm taking the derivative of this entire expression,\n

For the first term, I can just use the power\n

need to take down the exponent of two multiplied\n

Okay, so we're taking the derivative here\n

I'm taking the derivative of this entire expression,\n

For the first term, I can just use the power\n

need to take down the exponent of two multiplied\n

here, I do have my either the x function multiplied\n

the derivative of e to the x, which is either\nthe X

here, I do have my either the x function multiplied\n

the derivative of e to the x, which is either\nthe X

for my third part, I have just x times a constant\n

that constant. And so I just get e squared.\n

power of E squared, I can use the power rule\n

have a constant e squared in my exponent.\n

E squared times that by x and subtract one\n

fact that the derivative of e to the x is\njust e to

for my third part, I have just x times a constant\n

that constant. And so I just get e squared.\n

power of E squared, I can use the power rule\n

have a constant e squared in my exponent.\n

E squared times that by x and subtract one\n

fact that the derivative of e to the x is\njust e to

the x. This video prove some of the rules\n

the x. This video prove some of the rules\n

First, the constant row, it makes sense that\n

to be zero, because the slope of a horizontal\nline

First, the constant row, it makes sense that\n

to be zero, because the slope of a horizontal\nline

But we can also prove this fact, using the\n

of any function is the limit as h goes to\n

function at x divided by H. Well, here, our\n

the limit as h goes to zero of the constant\n

just the limit as h goes to zero of zero over\n

is zero. Intuitively, it also makes sense\n

x is got to be one, because the graph of y\n

But again, we can prove this using the limit\n

of x is the limit as h goes to zero of x plus\n

the limit of h over h, since the Xs can't\n

zero of one, which is one as wanted

But we can also prove this fact, using the\n

of any function is the limit as h goes to\nzero of

the function of x plus h minus the function\n

is just a constant. So we're taking the limit\n

constant divided by h, which is just the limit\n

just the limit of zero, which is zero. Intuitively,\n

the function y equals x is got to be one,\n

line with slope one. But again, we can prove\n

So the derivative of x is the limit as h goes\n

that simplifies to the limit of h over h,\n

limit as h goes to zero of one, which is one\nas wanted

h to the n minus one, and then finally, a\n

of x plus h to the n. Now, we still have to\n

and we still have to divide this whole thing\n

h to the n minus one, and then finally, a\n

of x plus h to the n. Now, we still have to\n

and we still have to divide this whole thing\n

notice that the X to the ends can So notice\n

in them. So if we factor out that H, we get\n

terms. And canceling the H's

notice that the X to the ends can So notice\n

in them. So if we factor out that H, we get\n

terms. And canceling the H's

one term that doesn't have any ages in it,\n

in them, as h goes to zero, all these other\n

what we're left with is simply n times x to\n

want for the power role. I think that's a\n

the binomial formula. But if you haven't seen\n

you feeling a little cold. So I'm going to\n

of the limit definition of derivative. So\n

over using this definition, f prime at a is\n

at x. So that's x to the n minus our function\n

a. Again, I'm going to need to rewrite things\n

currently in a zero over zero and determinant\n

out a copy of x minus a, which gives me x\n

A plus x to the n minus three A squared, you\n

I get to x a to the n minus two and finally\n

minus a, you can verify this factoring formula,\n

you In fact, do get x to the n minus a to\n

Now that I've factored, I can cancel my x\n

in x equal to A to get a to the n minus one\n

each of these terms is equal to a to the n\n

since we got them from the terms above that\n

with x to the zero. So that's n terms. So\n

a to the n minus one for a derivative f prime\n

Next, I'll prove the constant multiple rule\n

and f is a differentiable function, then the\n

constant times the derivative of f. Starting\n

have that the derivative of C times f of x\n

f of x plus h minus c times f of x over h.\n

of these terms, and actually I can pull it\n

constant has nothing to do with h. So now\n

the limit as h goes to zero of f of x plus\n

times the derivative of f, which is what we\nwanted to prove.

one term that doesn't have any ages in it,\n

in them, as h goes to zero, all these other\n

what we're left with is simply n times x to\n

want for the power role. I think that's a\n

the binomial formula. But if you haven't seen\n

you feeling a little cold. So I'm going to\n

of the limit definition of derivative. So\n

over using this definition, f prime at a is\n

at x. So that's x to the n minus our function\n

a. Again, I'm going to need to rewrite things\n

currently in a zero over zero and determinant\n

out a copy of x minus a, which gives me x\n

A plus x to the n minus three A squared, you\n

I get to x a to the n minus two and finally\n

minus a, you can verify this factoring formula,\n

you In fact, do get x to the n minus a to\n

Now that I've factored, I can cancel my x\n

in x equal to A to get a to the n minus one\n

each of these terms is equal to a to the n\n

since we got them from the terms above that\n

with x to the zero. So that's n terms. So\n

a to the n minus one for a derivative f prime\n

Next, I'll prove the constant multiple rule\n

and f is a differentiable function, then the\n

constant times the derivative of f. Starting\n

have that the derivative of C times f of x\n

f of x plus h minus c times f of x over h.\n

of these terms, and actually I can pull it\n

constant has nothing to do with h. So now\n

the limit as h goes to zero of f of x plus\n

times the derivative of f, which is what we\nwanted to prove.

The difference rule can be proved just like\n

of derivative and regrouping terms. Or we\n

together two of our previous roles. So if\n

of x plus minus one times g of x, then we\n

sum of derivatives, and then use the constant\n

one out, and then we have exactly what we\n

the proof of the content multiple rule, the\n

power rule when n is a positive integer. This\n

of functions that are products, or quotients\n

find statements of the product rule and the\n

proofs are in a separate video. Before we\n

rules. If f and g are differentiable functions,\n

g of x is just the sum of the derivatives.\n

The derivative of the difference is the difference\n

of rule hold for products of functions, in\n

equal to the product of the derivatives? Let's\n

example, if f of x is x, and g of x is x squared,\n

x times x squared, well, that's just the derivative\n

power rule. So it's 3x squared. On the other\n

we get one times 2x or just 2x. And these\n

the answer is no such a simple product rule\n

a product rule, it's just a little more complicated\n

rule says that \nif f and g are differentiable functions, then

the derivative of the product f of x times\n

of g of x plus the derivative of f of x times\n

of a product, we take the first function times\n

of the first times the second. Let's use this\n

square root of t times e to the t, we have\n

the derivative of the second function, plus\n

The difference rule can be proved just like\n

of derivative and regrouping terms. Or we\n

together two of our previous roles. So if\n

of x plus minus one times g of x, then we\n

sum of derivatives, and then use the constant\n

one out, and then we have exactly what we\n

the proof of the content multiple rule, the\n

power rule when n is a positive integer. This\n

of functions that are products, or quotients\n

find statements of the product rule and the\n

proofs are in a separate video. Before we\n

rules. If f and g are differentiable functions,\n

g of x is just the sum of the derivatives.\n

The derivative of the difference is the difference\n

of rule hold for products of functions, in\n

equal to the product of the derivatives? Let's\n

example, if f of x is x, and g of x is x squared,\n

x times x squared, well, that's just the derivative\n

power rule. So it's 3x squared. On the other\n

we get one times 2x or just 2x. And these\n

the answer is no such a simple product rule\n

a product rule, it's just a little more complicated\n

rule says that if f and g are differentiable\n

f of x times g of x is equal to f of x times\n

of f of x times g of x. In other words, to\n

the first function times the derivative of\n

times the second. Let's use this in an example.\n

of t times e to the t, we have to take the\n

of the second function, plus the derivative\n

times the second function. So that's the square\n

times the second function. So that's the square\n

And to find the derivative of the square root\n

in exponential form. Now we can just use the\n

one half minus one is negative one half and\n

this up a little bit, and I'm done. quotient\n

And to find the derivative of the square root\n

in exponential form. Now we can just use the\n

one half minus one is negative one half and\n

this up a little bit, and I'm done. quotient\n

is given By this quotient on the denominator,\n

And on the numerator, we have g of x times\nthe derivative

is given By this quotient on the denominator,\n

And on the numerator, we have g of x times\nthe derivative

minus f of x times the derivative of g of\n

If you think of f of x as the high function,\n

this is low D high minus high D low

minus f of x times the derivative of g of\n

If you think of f of x as the high function,\n

this is low D high minus high D low

we're low low means the low function squared.\n

So this derivative, taking us back to z here

we're low low means the low function squared.\n

So this derivative, taking us back to z here

we put the low low on the bottom, and then\nwe go low

we put the low low on the bottom, and then\nwe go low

z squared is to z minus high, D low the derivative\n

zero, don't really need to write the zero\n

z to the fourth plus two z minus three z to\nthe fourth

z squared is to z minus high, D low the derivative\n

zero, don't really need to write the zero\n

z to the fourth plus two z minus three z to\nthe fourth

not going to bother multiplying out this denominator,\n

So when I cancel things on the numerator,\n

z cubed plus one squared as the derivative\n

product rule and the quotient rule. I've written\n

of the dy dx notation, but you should check\n

For the proofs of these fabulously useful\n

In this video, I'll prove the product rule\n

related rule, called the reciprocal rule.\n

the derivative of the product, f of x times\n

limit definition of derivative. So the limit\n

plus h minus f of x g of x, Oliver H. Now\n

more like the expression above it, which is\n

going to use a classic trick of adding zero\n

So I'm going to rewrite my expression, leaving\n

as they are, but inserting two new terms that\n

x, g of x plus h, and then adding it back\n

expression. This is not as pointless as it\n

factor of g of x plus h from the first two\n

the next two terms. So I'm going to do that.\n

sum into two pieces here. You can use just\n

here and this expression here are the same.\n

expression here. Now my limit rules allow\n

limits, provided that these four limits do\n

in a moment that they do. So this first limit\n

a continuous function. G is continuous because\n

to be continuous. This second limit here,\n

derivative of f, so that limit exists at equals\n

x has nothing to do with age. So that limit\n

is the derivative of g. And we've done it.\n

see that this expression here, is exactly\n

the order of the terms switched around. Before\n

be really handy to prove the reciprocal rule,\n

one over f of x is given by negative the derivative\n

prove this fact, let's start as usual, with\n

of one over f of x is the limit as h goes\n

not going to bother multiplying out this denominator,\n

So when I cancel things on the numerator,\n

z cubed plus one squared as the derivative\n

product rule and the quotient rule. I've written\n

of the dy dx notation, but you should check\n

For the proofs of these fabulously useful\n

In this video, I'll prove the product rule\n

related rule, called the reciprocal rule.\n

the derivative of the product, f of x times\n

limit definition of derivative. So the limit\n

plus h minus f of x g of x, Oliver H. Now\n

more like the expression above it, which is\n

going to use a classic trick of adding zero\n

So I'm going to rewrite my expression, leaving\n

as they are, but inserting two new terms that\n

x, g of x plus h, and then adding it back\n

expression. This is not as pointless as it\n

factor of g of x plus h from the first two\n

the next two terms. So I'm going to do that.\n

sum into two pieces here. You can use just\n

here and this expression here are the same.\n

expression here. Now my limit rules allow\n

limits, provided that these four limits do\n

in a moment that they do. So this first limit\n

a continuous function. G is continuous because\n

to be continuous. This second limit here,\n

derivative of f, so that limit exists at equals\n

x has nothing to do with age. So that limit\n

is the derivative of g. And we've done it.\n

see that this expression here, is exactly\n

the order of the terms switched around. Before\n

be really handy to prove the reciprocal rule,\n

one over f of x is given by negative the derivative\n

prove this fact, let's start as usual, with\n

of one over f of x is the limit as h goes\n

h minus one over f of x over h. Now, these\n

by finding a common denominator that common\n

let me do that. I've just multiplied the first\n

fraction by f of x plus h over f of x plus\n

So that gives me f of x minus f of x plus\n

instead of dividing this whole thing by H,\n

gives me another factor of H in the denominator.\n

h minus one over f of x over h. Now, these\n

by finding a common denominator that common\n

let me do that. I've just multiplied the first\n

fraction by f of x plus h over f of x plus\n

So that gives me f of x minus f of x plus\n

instead of dividing this whole thing by H,\n

gives me another factor of H in the denominator.\n

It's just in the reverse order. So let me\n

me switch that order here. So this becomes\n

I've got the times one over f of x plus h\n

first, I'll factor out the negative sign.\n

of two limits, which I can do provided the\n

these limits do exist. Let's see the first\n

It's just in the reverse order. So let me\n

me switch that order here. So this becomes\n

I've got the times one over f of x plus h\n

first, I'll factor out the negative sign.\n

of two limits, which I can do provided the\n

these limits do exist. Let's see the first\n

And the second limit here exists because f\n

So by continuity, as H is going to zero, since\n

just approaching f of x. And I can rewrite\n

And so this is in other words, negative the\n

Now we've proved the reciprocal rule. Now\n

rule with very little effort. So instead of\n

this time, I'm just gonna think of the quotient\n

And the second limit here exists because f\n

So by continuity, as H is going to zero, since\n

just approaching f of x. And I can rewrite\n

And so this is in other words, negative the\n

Now we've proved the reciprocal rule. Now\n

rule with very little effort. So instead of\n

this time, I'm just gonna think of the quotient\n

And now, by the product rule, that's just\n

the second plus the derivative of the first\n

the derivative of this reciprocal is negative\n

still have this second term here, which I'm\n

And now, by the product rule, that's just\n

the second plus the derivative of the first\n

the derivative of this reciprocal is negative\n

still have this second term here, which I'm\n

so we're almost there. If we combine these\n

of g of x squared, we just have to multiply\n

to get that common denominator. Now we get\n

of x plus the derivative of f of x times g\n

this bottom expression is the same as this\n

the terms it is. So that's the end of the\n

gave proofs of the product rule, the reciprocal\n

is about two limits involving trig functions\n

limit as theta goes to zero of sine theta

so we're almost there. If we combine these\n

of g of x squared, we just have to multiply\n

to get that common denominator. Now we get\n

of x plus the derivative of f of x times g\n

this bottom expression is the same as this\n

the terms it is. So that's the end of the\n

gave proofs of the product rule, the reciprocal\n

is about two limits involving trig functions\n

limit as theta goes to zero of sine theta

over theta. And the limit as theta goes to\n

These limits turn out to have really nice\n

over theta. And the limit as theta goes to\n

These limits turn out to have really nice\n

not degrees. Let's consider the limit on the\n

to zero of sine theta over theta. Notice that\n

in zero for theta, because as theta goes to\n

to zero, and theta itself goes to zero, so\n

form. We can however, build up some evidence\n

calculator and a table of values, or by looking\n

here's the y axis. And you can see that as\n

the left, it's looking like the y value is\n

right, is also zero over zero and determinant\n

theta goes to one, so cosine theta minus one\n

we have some evidence to suggest that as theta\n

not degrees. Let's consider the limit on the\n

to zero of sine theta over theta. Notice that\n

in zero for theta, because as theta goes to\n

to zero, and theta itself goes to zero, so\n

form. We can however, build up some evidence\n

calculator and a table of values, or by looking\n

here's the y axis. And you can see that as\n

the left, it's looking like the y value is\n

right, is also zero over zero and determinant\n

theta goes to one, so cosine theta minus one\n

we have some evidence to suggest that as theta\n

these graphs provide strong evidence, but\n

for a rigorous proof. So for a pretty cool\n

please see the proof video for this section.\n

of sine theta over theta is one is really\n

Because intuitively, this is saying that sine\n

when theta is near zero, because the ratio\n

sine of this value of theta, without a calculator,\n

of 0.01769 is going to be approximately equal\n

you that when we're doing these limits, we're\n

not in radians, we won't get this nice limit\n

And we can check it on a calculator, and I\n

up to 10 decimal places. So as you can see,\n

use this same limit fat, again, in the next\n

as x goes to zero, the limit of tan of 7x\n

and signs and expression, I'm always tempted\n

and cosine, so I'm going to do that first\n

cosine. That still divided by sine of forex,\n

get sine of 7x over cosine of 7x times one\n

near zero, therefore 7x and 4x are also near\n

to 7x. And sine of 4x is approximately equal\n

pretty much the same thing as the limit as\n

4x. And canceling the access, this is just\n

of one of our cosine 7x. Since cosine of 7x\n

fourths. So this is the intuitive approach,\n

So more rigorously, I'm going to rewrite this\n

multiplying by 4x, over 4x, that hasn't changed\n

and fancy forms. But this is really useful.\n

7x over the 7x, times the one over cosine\n

the sign for x. And I'm still left with a\n

here, I can cancel out those x's. And I can\n

07, x is going to zero, so sine 7x over 7x\n

as x goes to zero, for x is going to zero.\n

reciprocal of one, it's also one. And finally,\n

07, x is going to zero, so cosine of 7x is\n

is going to one except the 7/4. So this limit\nis seven forth.

these graphs provide strong evidence, but\n

for a rigorous proof. So for a pretty cool\n

please see the proof video for this section.\n

of sine theta over theta is one is really\n

Because intuitively, this is saying that sine\n

when theta is near zero, because the ratio\n

sine of this value of theta, without a calculator,\n

of 0.01769 is going to be approximately equal\n

you that when we're doing these limits, we're\n

not in radians, we won't get this nice limit\n

And we can check it on a calculator, and I\n

up to 10 decimal places. So as you can see,\n

use this same limit fat, again, in the next\n

as x goes to zero, the limit of tan of 7x\n

and signs and expression, I'm always tempted\n

and cosine, so I'm going to do that first\n

cosine. That still divided by sine of forex,\n

get sine of 7x over cosine of 7x times one\n

near zero, therefore 7x and 4x are also near\n

to 7x. And sine of 4x is approximately equal\n

pretty much the same thing as the limit as\n

4x. And canceling the access, this is just\n

of one of our cosine 7x. Since cosine of 7x\n

fourths. So this is the intuitive approach,\n

So more rigorously, I'm going to rewrite this\n

multiplying by 4x, over 4x, that hasn't changed\n

and fancy forms. But this is really useful.\n

7x over the 7x, times the one over cosine\n

the sign for x. And I'm still left with a\n

here, I can cancel out those x's. And I can\n

07, x is going to zero, so sine 7x over 7x\n

as x goes to zero, for x is going to zero.\n

reciprocal of one, it's also one. And finally,\n

07, x is going to zero, so cosine of 7x is\n

is going to one except the 7/4. So this limit\nis seven forth.

In this video, we found that the limit as\n

is equal to one. And the limit as theta goes\n

is equal to zero. There's a nice proof of\n

When you compose two functions, you apply\n

second function to the output of the first\n

might compute population size from time in\nyears.

In this video, we found that the limit as\n

is equal to one. And the limit as theta goes\n

is equal to zero. There's a nice proof of\n

When you compose two functions, you apply\n

second function to the output of the first\n

might compute population size from time in\nyears.

So its input would be time in years, since\n

of people in the population. The second function\n

of population size. So it will take population\n

costs. If you put these functions together,\n

way from time in years to healthcare costs.\n

F. The composition of two functions, written\n

as follows. g composed with f of x is G evaluated\n

and so diagram, f acts on a number x and produces\n

f of x and produces a new number, g of f of\n

with F is the function that goes all the way\n

examples where our functions are defined by\n

with F of four, by definition, this means\n

we always work from the inside out. So we\n

f of four using the table of values for f\n

so we can replace F of four with the number\n

seven becomes our new x value in our table\n

to the G of X value of 10. So g of seven is\n

F of four is equal to 10. If instead we want\n

can rewrite that is f of g of four, and again,\n

to find g of four, so four is our x value.\n

that g of four is one. So we replaced by a\n

of one. Using our table for F values, f of\n

g of f of four, we got a different answer\n

in general, g composed with F is not the same\n

So its input would be time in years, since\n

of people in the population. The second function\n

of population size. So it will take population\n

costs. If you put these functions together,\n

way from time in years to healthcare costs.\n

F. The composition of two functions, written\n

as follows. g composed with f of x is G evaluated\n

and so diagram, f acts on a number x and produces\n

f of x and produces a new number, g of f of\n

with F is the function that goes all the way\n

examples where our functions are defined by\n

with F of four, by definition, this means\n

we always work from the inside out. So we\n

f of four using the table of values for f\n

so we can replace F of four with the number\n

seven becomes our new x value in our table\n

to the G of X value of 10. So g of seven is\n

F of four is equal to 10. If instead we want\n

can rewrite that is f of g of four, and again,\n

to find g of four, so four is our x value.\n

that g of four is one. So we replaced by a\n

of one. Using our table for F values, f of\n

g of f of four, we got a different answer\n

in general, g composed with F is not the same\n

pause the video and take a moment to compute\n

with F of two by the equivalent expression,\n

we know that f of two is three, and f of three\n

g of six, rewrite that as f of g of six isn't\n

of eight, eight is not on the table as an\n

there is no F of eight, this does not exist,\n

for F composed with g. Even though it was\n

the way through and get a value for F composed\n

to the composition of functions that are given\nby equations.

pause the video and take a moment to compute\n

with F of two by the equivalent expression,\n

we know that f of two is three, and f of three\n

g of six, rewrite that as f of g of six isn't\n

of eight, eight is not on the table as an\n

there is no F of eight, this does not exist,\n

for F composed with g. Even though it was\n

the way through and get a value for F composed\n

to the composition of functions that are given\nby equations.

p of x is x squared plus x and q of x is negative\n

p of x is x squared plus x and q of x is negative\n

As usual, I can rewrite this as Q of P of\n

is one squared plus one, so that's two. So\n

of two is negative two times two or negative\n

In this next example, we want to find q composed\n

as usual as Q of p of x and work from the\n

for that, that's x squared plus x. So I can\n

I'm stuck with evaluating q on x squared plus\n

times that thing. So q of x squared plus x\n

x squared plus x, what I've done is I've substituted\n

where I saw the X in this formula for q of\n

so that we'll be multiplying negative two\n

first piece, I can simplify this a bit as\n

expression for Q composed with p of x. Notice\n

P of one, which I already did in the first\n

now, negative two times one squared minus\n

before. Let's try another one. Let's try p\n

P of q of x. Working from the inside out,\n

I need to compute P of negative 2x. Here's\n

I need to plug in this expression everywhere\n

negative 2x squared plus negative 2x. Again,\n

I plug in the entire expression in for x.\n

2x. Notice that I got different expressions\n

again, we see that q composed with P is not\n

pause the video and try this last example\n

out, we're going to replace p of x with its\n

As usual, I can rewrite this as Q of P of\n

is one squared plus one, so that's two. So\n

of two is negative two times two or negative\n

In this next example, we want to find q composed\n

as usual as Q of p of x and work from the\n

for that, that's x squared plus x. So I can\n

I'm stuck with evaluating q on x squared plus\n

times that thing. So q of x squared plus x\n

x squared plus x, what I've done is I've substituted\n

where I saw the X in this formula for q of\n

so that we'll be multiplying negative two\n

first piece, I can simplify this a bit as\n

expression for Q composed with p of x. Notice\n

P of one, which I already did in the first\n

now, negative two times one squared minus\n

before. Let's try another one. Let's try p\n

P of q of x. Working from the inside out,\n

I need to compute P of negative 2x. Here's\n

I need to plug in this expression everywhere\n

negative 2x squared plus negative 2x. Again,\n

I plug in the entire expression in for x.\n

2x. Notice that I got different expressions\n

again, we see that q composed with P is not\n

pause the video and try this last example\n

out, we're going to replace p of x with its\n

everywhere we see an x in this formula, so\n

x squared plus x. Once again, I can simplify\n

fourth plus 2x cubed plus x squared plus x\n

cubed plus 2x squared plus x. In this last\n

we're given a formula for a function of h\n

as a composition of two functions, F and G.\n

functions gets applied first, f composed with\n

And since we evaluate these expressions from\n

and then F. In order to figure out what what\n

some thing inside my expression for H, so\n

seven, then whatever's inside the box, that'll\n

that gets applied, whatever happens to the\n

sign, that becomes my outside function, my\n

g of x is equal to x squared plus seven, and\n

just check and make sure that this works.\n

f composed with g, I need to get the same\n

do f composed with g of x, well, by definition,\n

everywhere we see an x in this formula, so\n

x squared plus x. Once again, I can simplify\n

fourth plus 2x cubed plus x squared plus x\n

cubed plus 2x squared plus x. In this last\n

we're given a formula for a function of h\n

as a composition of two functions, F and G.\n

functions gets applied first, f composed with\n

And since we evaluate these expressions from\n

and then F. In order to figure out what what\n

some thing inside my expression for H, so\n

seven, then whatever's inside the box, that'll\n

that gets applied, whatever happens to the\n

sign, that becomes my outside function, my\n

g of x is equal to x squared plus seven, and\n

just check and make sure that this works.\n

f composed with g, I need to get the same\n

do f composed with g of x, well, by definition,\n

Working from the inside out, I can replace\n

So I need to evaluate f of x squared plus\n

seven, into the formula for for F. So that\n

seven, two, it works because it matches my\n

a correct way of breaking h down as a composition\n

this is not the only correct answer. I'll\n

this time, I'll put the box in a different\n

that, then my inside function, my first function,\n

Working from the inside out, I can replace\n

So I need to evaluate f of x squared plus\n

seven, into the formula for for F. So that\n

seven, two, it works because it matches my\n

a correct way of breaking h down as a composition\n

this is not the only correct answer. I'll\n

this time, I'll put the box in a different\n

that, then my inside function, my first function,\n

So my f of x is what happens to the box, and\n

the square root. So in other words, f of x\n

Again, I can check that this works. If I do\n

So my f of x is what happens to the box, and\n

the square root. So in other words, f of x\n

Again, I can check that this works. If I do\n

So now g of x is x squared, so I'm taking\n

x, I do in fact get the square root of x squared\n

solution. In this video, we learn to evaluate\n

it and working from the inside out. We also\n

into a composition of two functions by boxing\n

first function applied in the composition.\n

second function applied in the composition\n

So now g of x is x squared, so I'm taking\n

x, I do in fact get the square root of x squared\n

solution. In this video, we learn to evaluate\n

it and working from the inside out. We also\n

into a composition of two functions by boxing\n

first function applied in the composition.\n

second function applied in the composition\n

This video is about solving rational equations.\n

that has rational expressions and that, in\n

in the denominator. There are several different\n

but they all start by finding the least common\n

are x plus three and x, we can think of one\n

the denominators don't have any factors in\n

just by multiplying them together. My next\n

By this, I mean that I multiply both sides\n

x plus three times x, I multiply on the left\n

same thing on the right side of the equation.\n

of the equation, I don't change the the value\n

denominator on both sides of the equation\n

terms in the equation, I can see this when\n

the same as before, pretty much. And then\n

three times x times one plus x plus three\n

multiplied the least common denominator by\n

have a blast canceling things. The x plus\n

denominator. The here are nothing cancels\n

are the x in the numerator cancels with the\n

expression as x squared equals x plus three\n

going to simplify. So I'll leave the x squared\n

squared plus 3x plus x plus three, hey, look,\n

so I get zero equals 4x plus three, so 4x\n

fourths. Finally, I'm going to plug in my\n

kind of equation. But it's especially important\n

for rational equations, you'll get what's\n

don't actually work in your original equation\n

in this example, I don't think we're going\n

three fourths is not going to make any of\n

out fine when I plug in. If I plug in, I get\n

negative three fourths plus three, three is\n

is one or flip and multiply to get minus four\n

fraction, it ends up being negative three\n

1/3. So that all seems to check out. And so\n

fourths. This next example looks a little\n

will work. First off, find the least common\n

c minus five, c plus one, and C squared minus\n

as C minus five times c plus one. Now, my\n

enough factors to that each of these denominators\n

five, I need the factor c plus one. And now\n

this denominator. So here is my least common\n

So I do this by multiplying both sides of\n

In fact, I can just multiply each of the three\n

went ahead and wrote my third denominator\n

what cancels. Now canceling time dies, this\n

This video is about solving rational equations.\n

that has rational expressions and that, in\n

in the denominator. There are several different\n

but they all start by finding the least common\n

are x plus three and x, we can think of one\n

the denominators don't have any factors in\n

just by multiplying them together. My next\n

By this, I mean that I multiply both sides\n

x plus three times x, I multiply on the left\n

same thing on the right side of the equation.\n

of the equation, I don't change the the value\n

denominator on both sides of the equation\n

terms in the equation, I can see this when\n

the same as before, pretty much. And then\n

three times x times one plus x plus three\n

multiplied the least common denominator by\n

have a blast canceling things. The x plus\n

denominator. The here are nothing cancels\n

are the x in the numerator cancels with the\n

expression as x squared equals x plus three\n

going to simplify. So I'll leave the x squared\n

squared plus 3x plus x plus three, hey, look,\n

so I get zero equals 4x plus three, so 4x\n

fourths. Finally, I'm going to plug in my\n

kind of equation. But it's especially important\n

for rational equations, you'll get what's\n

don't actually work in your original equation\n

in this example, I don't think we're going\n

three fourths is not going to make any of\n

out fine when I plug in. If I plug in, I get\n

negative three fourths plus three, three is\n

is one or flip and multiply to get minus four\n

fraction, it ends up being negative three\n

1/3. So that all seems to check out. And so\n

fourths. This next example looks a little\n

will work. First off, find the least common\n

c minus five, c plus one, and C squared minus\n

as C minus five times c plus one. Now, my\n

enough factors to that each of these denominators\n

five, I need the factor c plus one. And now\n

this denominator. So here is my least common\n

So I do this by multiplying both sides of\n

In fact, I can just multiply each of the three\n

went ahead and wrote my third denominator\n

what cancels. Now canceling time dies, this\n

cancel out the denominators, the whole point\n

you're multiplying by something that's big\n

you don't have to deal with denominators anymore.\n

So I get, let's see, c plus one times four\n

I get minus just c minus five, and then over\n

can rewrite the minus quantity c minus five\n

the three c squared from both sides to get\n

minus c, that becomes a three C. And finally,\n

get c squared plus three c plus two equals\n

looks like a nice one that factors. So this\n

zero. So either c plus one is zero, or C plus\n

C equals negative two. Now let's see, we need\n

to the trouble of calculating anything, I\n

going to work, because if I plug it in to\n

of zero, which doesn't make sense. So C equals\n

actually satisfy my original equation. And\n

negative two. I can go if I go ahead, and\n

So if I haven't made any mistakes, it should\n

just plug it in to be sure. And after some\n

final answer is C equals negative two. In\n

equations, using the method of finding the\n

the denominator, we cleared the denominator\n

by the least common denominator or equivalently.\n

There's another equivalent method that some\n

we find the least common denominator, but\n

least common denominator. So in this example,\n

cancel out the denominators, the whole point\n

you're multiplying by something that's big\n

you don't have to deal with denominators anymore.\n

So I get, let's see, c plus one times four\n

I get minus just c minus five, and then over\n

can rewrite the minus quantity c minus five\n

the three c squared from both sides to get\n

minus c, that becomes a three C. And finally,\n

get c squared plus three c plus two equals\n

looks like a nice one that factors. So this\n

zero. So either c plus one is zero, or C plus\n

C equals negative two. Now let's see, we need\n

to the trouble of calculating anything, I\n

going to work, because if I plug it in to\n

of zero, which doesn't make sense. So C equals\n

actually satisfy my original equation. And\n

negative two. I can go if I go ahead, and\n

So if I haven't made any mistakes, it should\n

just plug it in to be sure. And after some\n

final answer is C equals negative two. In\n

equations, using the method of finding the\n

the denominator, we cleared the denominator\n

by the least common denominator or equivalently.\n

There's another equivalent method that some\n

we find the least common denominator, but\n

least common denominator. So in this example,\n

But our next step would be to write each of\n

denominator by multiplying the top and the\n

in order to get the common denominator of\n

the bottom by x plus three times x, whenever\n

just by x plus three since that's what's missing\n

a little bit, let's say this is x squared\n

just x plus three times x over that denominator,\n

denominator. Now add together my fractions\n

So this is x plus three times x plus x plus\n

that are equal, that have the same denominator,\n

also. So the next step is to set the numerators\n

times x plus x plus three. And if you look\n

you'll recognize this equation. And so from\n

between these two methods, I personally tend\n

because it's a little bit less writing, you\n

earlier. You don't have to write them as many\n

bit easier to remember, a little easier to\n

One last caution. Don't forget at the end,\n

extraneous solutions. These will be solutions\n

equation. Go to zero. This video gives the\n

A graph of the function y equals sine x is\n

of the derivative of sine x by looking at\n

x equals zero, the tangent line has a positive\n

to pi over two, the slope of the tangent line\n

Next, the slope turns negative more and more\n

one, before returning again to zero. Continuing\n

y equals sine prime of x looks like the graph\n

video and do a similar exercise for the graph\n

the graph of y equals cosine x to estimate\n

prime of x. Notice that when x equals zero,\n

that slope turns negative, and then reaches\n

graph of the derivative should look something\n

the vertical reflection of the blue graph\n

of x is equal to the negative of sine of x.\n

of sine x is equal to cosine of x, and the\n

sine of x. For proofs of these facts, please\n

Once we have the derivatives of sine and cosine,\n

of a lot of other trig functions as well.

But our next step would be to write each of\n

denominator by multiplying the top and the\n

in order to get the common denominator of\n

the bottom by x plus three times x, whenever\n

just by x plus three since that's what's missing\n

a little bit, let's say this is x squared\n

just x plus three times x over that denominator,\n

denominator. Now add together my fractions\n

So this is x plus three times x plus x plus\n

that are equal, that have the same denominator,\n

also. So the next step is to set the numerators\n

times x plus x plus three. And if you look\n

you'll recognize this equation. And so from\n

between these two methods, I personally tend\n

because it's a little bit less writing, you\n

earlier. You don't have to write them as many\n

bit easier to remember, a little easier to\n

One last caution. Don't forget at the end,\n

extraneous solutions. These will be solutions\n

equation. Go to zero. This video gives the\n

A graph of the function y equals sine x is\n

of the derivative of sine x by looking at\n

x equals zero, the tangent line has a positive\n

to pi over two, the slope of the tangent line\n

Next, the slope turns negative more and more\n

one, before returning again to zero. Continuing\n

y equals sine prime of x looks like the graph\n

video and do a similar exercise for the graph\n

the graph of y equals cosine x to estimate\n

prime of x. Notice that when x equals zero,\n

that slope turns negative, and then reaches\n

graph of the derivative should look something\n

the vertical reflection of the blue graph\n

of x is equal to the negative of sine of x.\n

of sine x is equal to cosine of x, and the\n

sine of x. For proofs of these facts, please\n

Once we have the derivatives of sine and cosine,\n

of a lot of other trig functions as well.

And notice that a nice way to remember which\n

is that the derivatives of the trig functions\n

And notice that a nice way to remember which\n

is that the derivatives of the trig functions\n

always have a negative, and the root of the\n

positive. Now let's use these formulas in\n

involving several trig functions as well as\n

of how to proceed. I could try to rewrite\n

cosine and simplify, or I could attack the\n

I'm going to use the direct approach In this\n

will make things easier. So using the quotient\n

denominator squared. On the numerator, I get\n

cosine x, I need the product rule. So I get\n

negative sine x, plus the derivative of x,\n

have to do a minus Hi, x cosine of x dillow.\n

is a constant, plus the derivative of cotangent\n

I found the derivative, I'm going to go ahead\n

then rewriting everything in terms of sine\n

and denominator by sine squared of x, we have\n

you should memorize the derivatives of the\n

formulas are correct in a separate proof video.\n

special trig limit. And I'll also prove that\n

derivative of cosine is minus sign. To prove\n

one as theta goes to zero, I'm going to start\n

unit circle a circle of radius one, and I\n

and a smaller red triangle, both with angle\n

areas, if I want to compute the area of this\n

words, that pie shaped piece, I can first\n

times one squared for the radius. But since\n

has angle two pi, I need to multiply that\n

two pi to represent the fraction of the area\n

So in other words, the area of the sector\n

theta is given in the radians. Now if I want\n

I can do one half times the base times the\n

to cosine theta, because I have a circle of\n

is going to be sine theta. Finally, the area\n

the base times the height. But now the base\n

by tangent theta, since opposite, which is\n

has to equal tangent theta. Now if I put all\n

of the red triangle, alright is cosine theta\n

equal to the area of the blue sector, theta\n

area of the big green triangle, which is tan\n

through this inequality by two and rewrite\n

cosine theta sine theta is less than or equal\n

over cosine theta. Now I'm going to divide\n

won't change the inequalities as long as theta\n

always have a negative, and the root of the\n

positive. Now let's use these formulas in\n

involving several trig functions as well as\n

of how to proceed. I could try to rewrite\n

cosine and simplify, or I could attack the\n

I'm going to use the direct approach In this\n

will make things easier. So using the quotient\n

denominator squared. On the numerator, I get\n

cosine x, I need the product rule. So I get\n

negative sine x, plus the derivative of x,\n

have to do a minus Hi, x cosine of x dillow.\n

is a constant, plus the derivative of cotangent\n

I found the derivative, I'm going to go ahead\n

then rewriting everything in terms of sine\n

and denominator by sine squared of x, we have\n

you should memorize the derivatives of the\n

formulas are correct in a separate proof video.\n

special trig limit. And I'll also prove that\n

derivative of cosine is minus sign. To prove\n

one as theta goes to zero, I'm going to start\n

unit circle a circle of radius one, and I\n

and a smaller red triangle, both with angle\n

areas, if I want to compute the area of this\n

words, that pie shaped piece, I can first\n

times one squared for the radius. But since\n

has angle two pi, I need to multiply that\n

two pi to represent the fraction of the area\n

So in other words, the area of the sector\n

theta is given in the radians. Now if I want\n

I can do one half times the base times the\n

to cosine theta, because I have a circle of\n

is going to be sine theta. Finally, the area\n

the base times the height. But now the base\n

by tangent theta, since opposite, which is\n

has to equal tangent theta. Now if I put all\n

of the red triangle, alright is cosine theta\n

equal to the area of the blue sector, theta\n

area of the big green triangle, which is tan\n

through this inequality by two and rewrite\n

cosine theta sine theta is less than or equal\n

over cosine theta. Now I'm going to divide\n

won't change the inequalities as long as theta\n

And I get cosine theta is less than or equal\n

to one over cosine theta. Now this middle\n

I want to take the limit of. So I'm going\n

limits of the two expressions on the outside,\n

theorem, the limit of the expression on the\n

Now I've cheated a little bit here. And I've\n

because I've assumed that theta is greater\n

that less than zero, so that sign that as\n

equal one, the inequalities will flip around\n

theorem to get a limit of one. And that's\n

from calculus. To show that the limit of cosine\n

actually rewrite this expression and reuse\n

write down my limit. And I'm going to multiply\n

the numerator and the denominator. So I haven't\n

that by one. Now, if I multiply my numerator\nout

And I get cosine theta is less than or equal\n

to one over cosine theta. Now this middle\n

I want to take the limit of. So I'm going\n

limits of the two expressions on the outside,\n

theorem, the limit of the expression on the\n

Now I've cheated a little bit here. And I've\n

because I've assumed that theta is greater\n

that less than zero, so that sign that as\n

equal one, the inequalities will flip around\n

theorem to get a limit of one. And that's\n

from calculus. To show that the limit of cosine\n

actually rewrite this expression and reuse\n

write down my limit. And I'm going to multiply\n

the numerator and the denominator. So I haven't\n

that by one. Now, if I multiply my numerator\nout

I get cosine squared theta minus one. And\n

plus cosine squared theta equals one, I know\n

equal minus sine squared. So I can rewrite\n

theta over theta cosine theta plus one. And\n

theta, and my other copy of sine theta over\n

expression is going to be negative one, because\n

of the second expression is just zero over\n

limit is just going to be negative one times\n

it to prove. Now we can use these two limits\n

of sine and cosine, using the limit definition\n

stated previously. According to the limit\n

sine x is the limit as h goes to zero of sine\n

As usual, this is a zero for zero indeterminate\n

things to evaluate it. And I'm going to rewrite\n

sine of x plus h is equal to sine x cosine\n

things, and factor out a sine x from the first\n

and compute every piece. So this is sine x\n

my final answer is cosine x as we wanted the\n

sine is very similar. So please stop the video\n

Using the limit definition of derivative,\n

is the limit as h goes to zero of cosine of\n

rewrite the cosine of x plus h using the angle\n

of H minus the sine of x times the sine of\n

of x over h. As before, we're going to regroup\n

the first part, the same familiar limits just\n

of x as h goes to zero is just cosine of x.\n

saying sine of x, and sine of h over h is\n

is going to be negative sine of x times one\n

what we wanted. That's all for the proofs\n

motion or linear motion means the motion of\n

a particle moving left and right, or a ball\n

what the derivative and the second derivative\n

to move along a straight line. In this example,\n

line. And its position is given by this equation\n

particle is above its Baseline Position, whatever\n

mean the particle is below this Baseline Position.\n

prime of t. So by deriving I get four t cubed\n

derivative, and 12 t squared minus 32 t plus\n

I get cosine squared theta minus one. And\n

plus cosine squared theta equals one, I know\n

equal minus sine squared. So I can rewrite\n

theta over theta cosine theta plus one. And\n

theta, and my other copy of sine theta over\n

expression is going to be negative one, because\n

of the second expression is just zero over\n

limit is just going to be negative one times\n

it to prove. Now we can use these two limits\n

of sine and cosine, using the limit definition\n

stated previously. According to the limit\n

sine x is the limit as h goes to zero of sine\n

As usual, this is a zero for zero indeterminate\n

things to evaluate it. And I'm going to rewrite\n

sine of x plus h is equal to sine x cosine\n

things, and factor out a sine x from the first\n

and compute every piece. So this is sine x\n

my final answer is cosine x as we wanted the\n

sine is very similar. So please stop the video\n

Using the limit definition of derivative,\n

is the limit as h goes to zero of cosine of\n

rewrite the cosine of x plus h using the angle\n

of H minus the sine of x times the sine of\n

of x over h. As before, we're going to regroup\n

the first part, the same familiar limits just\n

of x as h goes to zero is just cosine of x.\n

saying sine of x, and sine of h over h is\n

is going to be negative sine of x times one\n

what we wanted. That's all for the proofs\n

motion or linear motion means the motion of\n

a particle moving left and right, or a ball\n

what the derivative and the second derivative\n

to move along a straight line. In this example,\n

line. And its position is given by this equation\n

particle is above its Baseline Position, whatever\n

mean the particle is below this Baseline Position.\n

prime of t. So by deriving I get four t cubed\n

derivative, and 12 t squared minus 32 t plus\n

S prime of t, which can also be written, D\n

of S of t, the position over time, well, the\n

And this can also be written as v of t, s\n

s with respect to t, can also be thought of\n

So that represents the rate of change of velocity\n

or decreasing. And that is called acceleration.\n

velocity and acceleration can be both positive\n

position is increasing. So the particle is\n

the position is decreasing, so the particle\n

means the particles at rest, at least for\n

equals mass times acceleration. So if the\n

the force is in the positive direction, it's\n

the other hand, the acceleration is negative\n

and it's like the particle is being pulled\n

no force on the particle at that instant,\n

these ideas about velocity acceleration. And\n

the particles motion at time equals 1.5 seconds.\n

is positive, so that means the particle is\n

is negative, so that means that its position\n

is moving down. Its acceleration is negative\n

So a negative acceleration means the velocity\n

decreasing is getting more and more negative.\n

and faster. This can be a little bit confusing,\n

it's getting more and more negative, the speed,\n

increasing. We can also see what the particle\n

graph, where the time is drawn on the x axis\n

the graph, we can see that when t is zero,\n

at its Baseline Position of zero. At time\n

position, but moving downwards. And since\n

and steeper, we can conclude that the speed\n

was we concluded from the table of values.\n

2.5 seconds. s of 2.5 seconds is negative.\n

Velocity s prime of t is also negative. So\n

the acceleration as double prime of t is positive.\n

well, a negative velocity that's increasing\n

the particle must be slowing down. And in\n

graph agrees with this reasoning, at 2.5 seconds,\n

decreasing, so the particles moving down,\n

the particle speed is decreasing. Even though\nit's velocity

S prime of t, which can also be written, D\n

of S of t, the position over time, well, the\n

And this can also be written as v of t, s\n

s with respect to t, can also be thought of\n

So that represents the rate of change of velocity\n

or decreasing. And that is called acceleration.\n

velocity and acceleration can be both positive\n

position is increasing. So the particle is\n

the position is decreasing, so the particle\n

means the particles at rest, at least for\n

equals mass times acceleration. So if the\n

the force is in the positive direction, it's\n

the other hand, the acceleration is negative\n

and it's like the particle is being pulled\n

no force on the particle at that instant,\n

these ideas about velocity acceleration. And\n

the particles motion at time equals 1.5 seconds.\n

is positive, so that means the particle is\n

is negative, so that means that its position\n

is moving down. Its acceleration is negative\n

So a negative acceleration means the velocity\n

decreasing is getting more and more negative.\n

and faster. This can be a little bit confusing,\n

it's getting more and more negative, the speed,\n

increasing. We can also see what the particle\n

graph, where the time is drawn on the x axis\n

the graph, we can see that when t is zero,\n

at its Baseline Position of zero. At time\n

position, but moving downwards. And since\n

and steeper, we can conclude that the speed\n

was we concluded from the table of values.\n

2.5 seconds. s of 2.5 seconds is negative.\n

Velocity s prime of t is also negative. So\n

the acceleration as double prime of t is positive.\n

well, a negative velocity that's increasing\n

the particle must be slowing down. And in\n

graph agrees with this reasoning, at 2.5 seconds,\n

decreasing, so the particles moving down,\n

the particle speed is decreasing. Even though\nit's velocity

which you can think of as speed with direction\n

velocity that's getting less negative. Notice\n

acceleration are both in the same direction,\n

was speeding up. And the second example, were\n

directions one positive one negative, the\n

which you can think of as speed with direction\n

velocity that's getting less negative. Notice\n

acceleration are both in the same direction,\n

was speeding up. And the second example, were\n

directions one positive one negative, the\n

This is true in general, when velocity acceleration\n

or both negative, then the particle is speeding\nup.

This is true in general, when velocity acceleration\n

or both negative, then the particle is speeding\nup.

And when velocity acceleration have opposite\n

One way to think about this is in terms of\n

So if velocity acceleration have the same\n

as the particles already going, so it's making\n

acceleration have opposite signs, then the\n

moving, so it's causing it to slow down. Let's\n

it'll be helpful to write down the velocity\n

earlier. I've also graphed position, velocity\n

you go on, it's a fun exercise to figure out\n

the equations just based on the shapes of\n

they're decreasing where they're positive\n

derivative of position. So velocity needs\n

The only pairs of functions that have this\n

when the red ones increasing, and the green\n

one is increasing. Now acceleration, which\n

be positive, when velocity is increasing.\n

with both of these relationships is to make\n

and the green one be acceleration. This agrees\n

And when velocity acceleration have opposite\n

One way to think about this is in terms of\n

So if velocity acceleration have the same\n

as the particles already going, so it's making\n

acceleration have opposite signs, then the\n

moving, so it's causing it to slow down. Let's\n

it'll be helpful to write down the velocity\n

earlier. I've also graphed position, velocity\n

you go on, it's a fun exercise to figure out\n

the equations just based on the shapes of\n

they're decreasing where they're positive\n

derivative of position. So velocity needs\n

The only pairs of functions that have this\n

when the red ones increasing, and the green\n

one is increasing. Now acceleration, which\n

be positive, when velocity is increasing.\n

with both of these relationships is to make\n

and the green one be acceleration. This agrees\n

The first question asks, When is the particle\n

when the velocity is zero. In other words,\n

for S prime of t, we can factor out a four\n

The first question asks, When is the particle\n

when the velocity is zero. In other words,\n

for S prime of t, we can factor out a four\n

This conclusion agrees with our graph of V\n

and also agrees with our graph of position\n

it changed direction, when t equals 01. And\n

is positive, and moving down when velocity\n

question, the velocity equals zero, when t\n

those values to figure out whether the velocity\n

in values. So for example, when t is negative\n

equation for velocity, I get a negative number.\n

than zero, between zero and one. If I plug\n

a value of S prime of t or V of t of 2.5,\n

value of t in between one and three, say t\n

eight, which is a negative number. And finally,\n

so For, I get a positive answer for V of t.\n

is negative when t is between negative infinity\n

V of t is positive when t is between

This conclusion agrees with our graph of V\n

and also agrees with our graph of position\n

it changed direction, when t equals 01. And\n

is positive, and moving down when velocity\n

question, the velocity equals zero, when t\n

those values to figure out whether the velocity\n

in values. So for example, when t is negative\n

equation for velocity, I get a negative number.\n

than zero, between zero and one. If I plug\n

a value of S prime of t or V of t of 2.5,\n

value of t in between one and three, say t\n

eight, which is a negative number. And finally,\n

so For, I get a positive answer for V of t.\n

is negative when t is between negative infinity\n

V of t is positive when t is between

one, and between three and infinity. Of course,\n

by looking at the graph of velocity and where\n

looking at the graph of position and seeing\n

To answer the next question, the particle\n

are both positive or both negative. And the\n

and a of t have opposite signs. So let's make\n

of t is positive and negative. First, it'll\n

So if I set zero equal to my S double prime,\n

one, and between three and infinity. Of course,\n

by looking at the graph of velocity and where\n

looking at the graph of position and seeing\n

To answer the next question, the particle\n

are both positive or both negative. And the\n

and a of t have opposite signs. So let's make\n

of t is positive and negative. First, it'll\n

So if I set zero equal to my S double prime,\n

and then use the quadratic equation to find\n

factor easily, this simplifies to four thirds\n

three, which is approximately 0.45, and 2.22.\n

mark the places where acceleration is zero.\n

is the equation for acceleration, I get a\n

t equals one, I get a negative answer here.\n

get another positive answer here. Now, if\n

which changed sign at 01, and three, and went\n

positive, I can try to figure out where velocity\n

be helpful actually to shade in where acceleration\n

velocity is positive. And then look for the\n

and 4.5, and greater than three. And then\n

looks like in between one and 2.22. So that's\n

know that V of t and a of t have opposite\n

we'll use exact values of four thirds plus\n

these decimal approximations. So let me write\n

up. And here, it's where it's slowing down,\n

of position, the particle speeding up with\n

steeper, that's the red graph is getting steeper\n

we found algebraically. As our final example,\n

traveled between one and four seconds for\n

is five thirds, or about 1.67 millimeters\n

or about 10.67 millimeters, all I'm doing\n

So the net change in position is just the\n

minus as of one, which is nine millimeters.\n

distance traveled between one and four seconds\n

it's a little more complicated. Because the\n

period, it doesn't go straight from its position\n

Remember what the graph Position looked like\n

and at three seconds. So to find the total\n

from one second to three seconds, plus the\n

four seconds. Another way of thinking about\n

s three minus s one, plus the absolute value\n

value signs because this difference in position\n

the particles moving down. Plugging in the\n

27 minus five thirds, plus the absolute value\n

total of 199 thirds, or 66.3 repeating millimeters,\n

difference in position. This video gave an\n

and then use the quadratic equation to find\n

factor easily, this simplifies to four thirds\n

three, which is approximately 0.45, and 2.22.\n

mark the places where acceleration is zero.\n

is the equation for acceleration, I get a\n

t equals one, I get a negative answer here.\n

get another positive answer here. Now, if\n

which changed sign at 01, and three, and went\n

positive, I can try to figure out where velocity\n

be helpful actually to shade in where acceleration\n

velocity is positive. And then look for the\n

and 4.5, and greater than three. And then\n

looks like in between one and 2.22. So that's\n

know that V of t and a of t have opposite\n

we'll use exact values of four thirds plus\n

these decimal approximations. So let me write\n

up. And here, it's where it's slowing down,\n

of position, the particle speeding up with\n

steeper, that's the red graph is getting steeper\n

we found algebraically. As our final example,\n

traveled between one and four seconds for\n

is five thirds, or about 1.67 millimeters\n

or about 10.67 millimeters, all I'm doing\n

So the net change in position is just the\n

minus as of one, which is nine millimeters.\n

distance traveled between one and four seconds\n

it's a little more complicated. Because the\n

period, it doesn't go straight from its position\n

Remember what the graph Position looked like\n

and at three seconds. So to find the total\n

from one second to three seconds, plus the\n

four seconds. Another way of thinking about\n

s three minus s one, plus the absolute value\n

value signs because this difference in position\n

the particles moving down. Plugging in the\n

27 minus five thirds, plus the absolute value\n

total of 199 thirds, or 66.3 repeating millimeters,\n

difference in position. This video gave an\n

A similar analysis could be done for a particle\n

where a positive position means the particles\n

means the particles on the left side of his\n

can be done for other objects, not just particles.\n

straight up and then falling down again. This\n

the derivative to a cost function. Suppose\n

A similar analysis could be done for a particle\n

where a positive position means the particles\n

means the particles on the left side of his\n

can be done for other objects, not just particles.\n

straight up and then falling down again. This\n

the derivative to a cost function. Suppose\n

I'm going to sketch a few graphs, and you\n

representation for C of x. Pause the video\n

candidate graphs that I've drawn have a nonzero\n

that there's some fixed startup cost and buying\n

Now, I would like to suggest that C of x should\n

going to cost more money to make more t shirts,\n

function is out. Now it's somewhat reasonable,\n

of x like it is here, if you've got the same\n

or 1000 t shirts, the slope in that case would\n

function would mean that cost per t shirt\n

making. But in reality, it's probably going\n

is to make just a few t shirts. And therefore\n

going down as x increases. So this function\n

down for larger access. And so I would say\n

for C of x as a function of x. In other words,\n

C prime of x should be decreasing. C of 204\n

for making 204 t shirts instead of 200. In\n

of making the last four t shirts. The ratio\n

the average rate of change of C of x. The\n

dollars per t shirt. And formula you might\n

shirt of making the last four t shirts. C\n

change of C of x. c of x is known as the cost\n

cost, which is the rate at which cost is increasing\n

a little bit weird to take the derivative\n

integer values. But we can always approximate\n

real numbers. To make this a little more specific,\n

500 plus 300 times the square root of x. In\n

of iPads that are produced, and C of x is\n

C of 401 minus C of 400. given by 500 plus\n

plus 300 times the square root of 400. This\n

cent. This means that it costs an additional\n

iPad. In this fictitious example

I'm going to sketch a few graphs, and you\n

representation for C of x. Pause the video\n

candidate graphs that I've drawn have a nonzero\n

that there's some fixed startup cost and buying\n

Now, I would like to suggest that C of x should\n

going to cost more money to make more t shirts,\n

function is out. Now it's somewhat reasonable,\n

of x like it is here, if you've got the same\n

or 1000 t shirts, the slope in that case would\n

function would mean that cost per t shirt\n

making. But in reality, it's probably going\n

is to make just a few t shirts. And therefore\n

going down as x increases. So this function\n

down for larger access. And so I would say\n

for C of x as a function of x. In other words,\n

C prime of x should be decreasing. C of 204\n

for making 204 t shirts instead of 200. In\n

of making the last four t shirts. The ratio\n

the average rate of change of C of x. The\n

dollars per t shirt. And formula you might\n

shirt of making the last four t shirts. C\n

change of C of x. c of x is known as the cost\n

cost, which is the rate at which cost is increasing\n

a little bit weird to take the derivative\n

integer values. But we can always approximate\n

real numbers. To make this a little more specific,\n

500 plus 300 times the square root of x. In\n

of iPads that are produced, and C of x is\n

C of 401 minus C of 400. given by 500 plus\n

plus 300 times the square root of 400. This\n

cent. This means that it costs an additional\n

iPad. In this fictitious example

if I want to compute C prime of 400, instead,\n

times one half x to the minus one half. So\n

half times 400 to the negative one half, which\n

root of 400, which is also 7.5, or $7.50.\n

these two answers are equal. And it makes\n

this difference. Since C prime of 400, the\n

rate of change going from 400 to 401, which\n

again, C prime of 400 is called the marginal\n

cost function is increasing with each additional\n

function, and it's derivative, which is known\n

logarithms are a way of writing exponents.\n

that a to the C equals b. In other words,\n

a to to get be. The number A is called the\n

base when we write it in this exponential\n

this relationship, log base a of B equals\n

if I want to compute C prime of 400, instead,\n

times one half x to the minus one half. So\n

half times 400 to the negative one half, which\n

root of 400, which is also 7.5, or $7.50.\n

these two answers are equal. And it makes\n

this difference. Since C prime of 400, the\n

rate of change going from 400 to 401, which\n

again, C prime of 400 is called the marginal\n

cost function is increasing with each additional\n

function, and it's derivative, which is known\n

logarithms are a way of writing exponents.\n

that a to the C equals b. In other words,\n

a to to get be. The number A is called the\n

base when we write it in this exponential\n

this relationship, log base a of B equals\n

Other students like to think of it in terms\n

What power do you raise a two in order to\n

two of eight is three, because two to the\n

of y is asking you the question, What power\n

example, log base two of 16 is four, because\n

equals 16? And the answer is four. Please\n

examples. log base two of two is asking, What\n

answer is one. Two to the one equals two.\n

what power gives you one half? Well, to get\n

power. So that would be two to the negative\n

two of 1/8 means what power do we raise to\n

one over two cubed, we have to raise two to\n

cubed. So our exponent is negative three,\n

Finally, log base two of one is asking to\n

to the zero power gives us one, so this log\n

we can get positive negative and zero answers\n

the video and figure out what these logs evaluate\n

notice that a million is 10 to the sixth power.\n

do we raise tend to to get a million? So that\n

to the six? Well, of course, the answer is\n

one is 10 to the minus three, this log expressions,\n

10 of 10 to the minus three? Well, what power\n

minus three? Of course, the answer is negative\n

power do we raise 10 to to get zero. If you\n

to an exponent get zero. Raising 10 to a positive\n

Raising 10 to a negative exponent is like\n

fractions, but they're still positive numbers,\n

raised 10 to the zero power, we'll just get\n

log base 10 of zero does not exist. If you\n

10 button, you'll get an error message. Same\n

100. We're asking 10 to what power equals\n

will work. And more generally, it's possible\n

than zero, but not for numbers that are less\n

domain of the function log base a of x, no\n

domain is going to be all positive numbers.\n

x, that's called natural log, and it means\n

number that's about 2.718. When you see log\n

means log base 10 of x, and it's called the\n

buttons for natural log, and for common log.\n

logs in them. log base three of one nine is\n

three to the negative two equals 1/9. Log\n

So that can be rewritten as 10 to the 1.11394\n

ln means natural log, or log base e, so I\n

whenever E equals negative one. Well, that\n

one equals one over e, which is true. Now\n

with exponential equations and rewrite them\n

Other students like to think of it in terms\n

What power do you raise a two in order to\n

two of eight is three, because two to the\n

of y is asking you the question, What power\n

example, log base two of 16 is four, because\n

equals 16? And the answer is four. Please\n

examples. log base two of two is asking, What\n

answer is one. Two to the one equals two.\n

what power gives you one half? Well, to get\n

power. So that would be two to the negative\n

two of 1/8 means what power do we raise to\n

one over two cubed, we have to raise two to\n

cubed. So our exponent is negative three,\n

Finally, log base two of one is asking to\n

to the zero power gives us one, so this log\n

we can get positive negative and zero answers\n

the video and figure out what these logs evaluate\n

notice that a million is 10 to the sixth power.\n

do we raise tend to to get a million? So that\n

to the six? Well, of course, the answer is\n

one is 10 to the minus three, this log expressions,\n

10 of 10 to the minus three? Well, what power\n

minus three? Of course, the answer is negative\n

power do we raise 10 to to get zero. If you\n

to an exponent get zero. Raising 10 to a positive\n

Raising 10 to a negative exponent is like\n

fractions, but they're still positive numbers,\n

raised 10 to the zero power, we'll just get\n

log base 10 of zero does not exist. If you\n

10 button, you'll get an error message. Same\n

100. We're asking 10 to what power equals\n

will work. And more generally, it's possible\n

than zero, but not for numbers that are less\n

domain of the function log base a of x, no\n

domain is going to be all positive numbers.\n

x, that's called natural log, and it means\n

number that's about 2.718. When you see log\n

means log base 10 of x, and it's called the\n

buttons for natural log, and for common log.\n

logs in them. log base three of one nine is\n

three to the negative two equals 1/9. Log\n

So that can be rewritten as 10 to the 1.11394\n

ln means natural log, or log base e, so I\n

whenever E equals negative one. Well, that\n

one equals one over e, which is true. Now\n

with exponential equations and rewrite them\n

the base stays the same in both expressions.\n

the exponential equation, that's going to\n

just have to figure out what's in the argument\n

of the equal sign. Remember that the answer\n

goes in this box should be my exponent for\n

and I'll put the 9.78 as the argument of my\n

9.78 equals u means the same thing as three\n

started with. In the second example, the base\n

of my log is going to be the answer to my\n

3x plus seven. And the other expression, the\n

Let me check, log base e of four minus y equals\n

four minus Y, which is just what I started\n

log. This video introduced the idea of logs.\n

means the same thing as a to the C equals\n

What power exponent Do you raise a to in order\n

graph, so some log functions and also talk\n

let's graph a log function by hand by plotting\n

is y equals log base two of x, I'll make a\n

this out by hand, I want to pick x values\n

of x. So I'll start out with the x value of\n

log base anything of one is 02 is another\n

of two, that's asking, What power do I raise\n

Power other powers of two are easy to work\n

that saying what power do I raise to to to\n

log base two of eight is three, and log base\n

fractional values for X. If x is one half,\n

what power do I raise to two to get one half?\n

It's also easy to compute by hand, the log\n

is negative two, since two to the negative\n

1/8 is negative three. I'll put some tick\n

video and take a moment to plot these points.\n

here to one that's here, for two, that is\n

the base stays the same in both expressions.\n

the exponential equation, that's going to\n

just have to figure out what's in the argument\n

of the equal sign. Remember that the answer\n

goes in this box should be my exponent for\n

and I'll put the 9.78 as the argument of my\n

9.78 equals u means the same thing as three\n

started with. In the second example, the base\n

of my log is going to be the answer to my\n

3x plus seven. And the other expression, the\n

Let me check, log base e of four minus y equals\n

four minus Y, which is just what I started\n

log. This video introduced the idea of logs.\n

means the same thing as a to the C equals\n

What power exponent Do you raise a to in order\n

graph, so some log functions and also talk\n

let's graph a log function by hand by plotting\n

is y equals log base two of x, I'll make a\n

this out by hand, I want to pick x values\n

of x. So I'll start out with the x value of\n

log base anything of one is 02 is another\n

of two, that's asking, What power do I raise\n

Power other powers of two are easy to work\n

that saying what power do I raise to to to\n

log base two of eight is three, and log base\n

fractional values for X. If x is one half,\n

what power do I raise to two to get one half?\n

It's also easy to compute by hand, the log\n

is negative two, since two to the negative\n

1/8 is negative three. I'll put some tick\n

video and take a moment to plot these points.\n

here to one that's here, for two, that is\n

And the fractional x values, one half goes\n

And the fractional x values, one half goes\n

And if I connect the dots, I get a graph that\n

and smaller fractions, I would keep getting\n

log base two of them, so my graph is getting\n

more and more negative as x is getting close\n

graph over here with negative X values, I\n

That omission is no accident. Because if you\n

of a negative number, like say negative four\n

exist because there's no power that you can\n

there are no points on the graph for negative\n

on the graph where x is zero, because you\n

power you can raise to to to get zero. I want\n

First of all, the domain is x values greater\n

that as a round bracket because I don't want\n

going to be the y values, while they go all\n

numbers. And the graph gradually increases\n

range is actually all real numbers are an\n

Finally, I want to point out that this graph\n

is at the line x equals zero. I'll draw that\n

asymptote is a line that our functions graph\n

of y equals log base two of x. But if I wanted\n

would look very similar, it would still have\n

range of all real numbers and a vertical asymptote\n

point one zero, but it would go through the\n

of 10 is one, it would look pretty much the\n

though it doesn't look like it with the way\n

to n towards infinity. In fact, the graph\n

than one looks pretty much the same, and has\n

what the basic log graph looks like, we can\n

without plotting points. Here we have the\n

I'm just going to draw a rough graph. If I\n

would want to plot some points. But I know\n

y equals ln of x, that would look something\n

And if I connect the dots, I get a graph that\n

and smaller fractions, I would keep getting\n

log base two of them, so my graph is getting\n

more and more negative as x is getting close\n

graph over here with negative X values, I\n

That omission is no accident. Because if you\n

of a negative number, like say negative four\n

exist because there's no power that you can\n

there are no points on the graph for negative\n

on the graph where x is zero, because you\n

power you can raise to to to get zero. I want\n

First of all, the domain is x values greater\n

that as a round bracket because I don't want\n

going to be the y values, while they go all\n

numbers. And the graph gradually increases\n

range is actually all real numbers are an\n

Finally, I want to point out that this graph\n

is at the line x equals zero. I'll draw that\n

asymptote is a line that our functions graph\n

of y equals log base two of x. But if I wanted\n

would look very similar, it would still have\n

range of all real numbers and a vertical asymptote\n

point one zero, but it would go through the\n

of 10 is one, it would look pretty much the\n

though it doesn't look like it with the way\n

to n towards infinity. In fact, the graph\n

than one looks pretty much the same, and has\n

what the basic log graph looks like, we can\n

without plotting points. Here we have the\n

I'm just going to draw a rough graph. If I\n

would want to plot some points. But I know\n

y equals ln of x, that would look something\n

with a vertical asymptote along the y axis.\n

that just shifts our graph by five units,\n

since the vertical line shifted up by five\n

of going through one zero, it'll go through\n

sketch here. Let's compare our starting function\n

y equals ln x plus five in terms of the domain,\n

original function y equals ln x had a domain\n

the outside affects the y values, and the\n

doesn't change the domain. So the domain is\n

of our original y equals ln x was from negative\n

does affect the y values, and the range is\n

original range was all real numbers, if you\n

still get the set of all real numbers. So\n

And finally, we already saw that the original\n

zero, when we shift that up by five units,\n

In this next example, we're starting with\n

two is on the inside, that means we shift\n

log function. Here's our basic log function.\n

going through the point one, zero, here's\n

everything left by two. So my vertical asymptote\n

negative two, instead of at x equals zero,\n

gets shifted to, let's see negative one zero,\n

here's a rough sketch of the resulting graph.\n

drawn here. We're talking about domains, the\n

But now I've shifted that left. So I subtracted\n

domain, which I can also verify just by looking\n

negative infinity to infinity. Well, shifting\n

even affect the range. So my range is still\n

asymptote was originally at x equals zero.\n

that shifts that to x equals negative two.\n

about drawing this graph. I'll just use algebra\n

What's the issue, when you're taking the logs\n

a negative number or zero. So whatever is\n

is being fed into log had better be greater\n

to minus 3x, to be greater than zero. Now\n

has got to be greater than 3x. So two thirds\n

be less than two thirds. So our domain is\n

two thirds, not including two thirds, it's\n

the graph of a log function. It looks something\n

and has a vertical asymptote on the y axis.\n

the log of a negative number, or zero, then\n

log functions. Whatever's inside the log function,\n

This video is about combining logs and exponents.\n

use your calculator to evaluate the following\nfour expressions.

with a vertical asymptote along the y axis.\n

that just shifts our graph by five units,\n

since the vertical line shifted up by five\n

of going through one zero, it'll go through\n

sketch here. Let's compare our starting function\n

y equals ln x plus five in terms of the domain,\n

original function y equals ln x had a domain\n

the outside affects the y values, and the\n

doesn't change the domain. So the domain is\n

of our original y equals ln x was from negative\n

does affect the y values, and the range is\n

original range was all real numbers, if you\n

still get the set of all real numbers. So\n

And finally, we already saw that the original\n

zero, when we shift that up by five units,\n

In this next example, we're starting with\n

two is on the inside, that means we shift\n

log function. Here's our basic log function.\n

going through the point one, zero, here's\n

everything left by two. So my vertical asymptote\n

negative two, instead of at x equals zero,\n

gets shifted to, let's see negative one zero,\n

here's a rough sketch of the resulting graph.\n

drawn here. We're talking about domains, the\n

But now I've shifted that left. So I subtracted\n

domain, which I can also verify just by looking\n

negative infinity to infinity. Well, shifting\n

even affect the range. So my range is still\n

asymptote was originally at x equals zero.\n

that shifts that to x equals negative two.\n

about drawing this graph. I'll just use algebra\n

What's the issue, when you're taking the logs\n

a negative number or zero. So whatever is\n

is being fed into log had better be greater\n

to minus 3x, to be greater than zero. Now\n

has got to be greater than 3x. So two thirds\n

be less than two thirds. So our domain is\n

two thirds, not including two thirds, it's\n

the graph of a log function. It looks something\n

and has a vertical asymptote on the y axis.\n

the log of a negative number, or zero, then\n

log functions. Whatever's inside the log function,\n

This video is about combining logs and exponents.\n

use your calculator to evaluate the following\nfour expressions.

Remember, that log base 10 on your calculator\n

calculator is the natural log button, you\n

is three, the log base e of e to the 4.2 is\n

And eat the log base e of 9.6 is 9.6. In each\n

with the same base undo each other, and we're\n

that for any base a the log base a of a to\n

happens if we do the exponential function\n

the opposite order. For example, we take 10\n

to the power and the log base 10 undo each\n

happens for any base a a to the log base a\n

saying that an exponential function and a\n

other. If you're familiar with the language\n

and log function are inverses. Let's see why\n

base a of a dx is asking the question, What\n

to the x? In other words, a to what power\n

x. And that's why log base a of a to the x\n

that the log base a of x means the power we\n

is saying that we're supposed to raise a to\n

need to raise a two to get x, then we'll certainly\n

examples. If we want to find three to the\n

base three undo each other, so we're left\nwith 1.4.

Remember, that log base 10 on your calculator\n

calculator is the natural log button, you\n

is three, the log base e of e to the 4.2 is\n

And eat the log base e of 9.6 is 9.6. In each\n

with the same base undo each other, and we're\n

that for any base a the log base a of a to\n

happens if we do the exponential function\n

the opposite order. For example, we take 10\n

to the power and the log base 10 undo each\n

happens for any base a a to the log base a\n

saying that an exponential function and a\n

other. If you're familiar with the language\n

and log function are inverses. Let's see why\n

base a of a dx is asking the question, What\n

to the x? In other words, a to what power\n

x. And that's why log base a of a to the x\n

that the log base a of x means the power we\n

is saying that we're supposed to raise a to\n

need to raise a two to get x, then we'll certainly\n

examples. If we want to find three to the\n

base three undo each other, so we're left\nwith 1.4.

If we want to find ln of e to the x, remember\n

If we want to find ln of e to the x, remember\n

Well, those functions undo each other and\n

the log of three z, remember that log without\n

So really, we want to take 10 to the log base\n

base 10 undo each other. So we're left with\n

hold is ln of 10 to the x equal to x? Well,\n

e of 10 to the x, notice that the base of\n

are not the same thing. So they don't undo\n

to the x is not usually equal to x, we can\n

then log base e of 10 to the one, that's log\n

that's equal to 2.3. In some more decimals,\n

statement is false, it does not hold. We need\n

to undo each other. In this video, we saw\n

undo each other. Specifically, log base a\n

base a of x is also equal to x

Well, those functions undo each other and\n

the log of three z, remember that log without\n

So really, we want to take 10 to the log base\n

base 10 undo each other. So we're left with\n

hold is ln of 10 to the x equal to x? Well,\n

e of 10 to the x, notice that the base of\n

are not the same thing. So they don't undo\n

to the x is not usually equal to x, we can\n

then log base e of 10 to the one, that's log\n

that's equal to 2.3. In some more decimals,\n

statement is false, it does not hold. We need\n

to undo each other. In this video, we saw\n

undo each other. Specifically, log base a\n

base a of x is also equal to x

for any values of x and any base a. This video\n

log rules are closely related to the exponent\n

for any values of x and any base a. This video\n

log rules are closely related to the exponent\n

To keep things simple, we'll write everything\n

rules hold for any base, we know that if we\n

have a product rule for exponents, which says\n

to two to the m plus n. In other words, if\n

We also have a quotient rule that says that\n

to two to the m minus n. In words, that says\n

the exponents. Finally, we have a power rule\n

we multiply the exponents. Each of these exponent\n

first rule, two to the zero equals one can\n

two of one equals zero. That's because log\n

zero equals one. The second rule, the product\n

saying log of x times y equals log of x plus\n

with my base that I'm using for my exponent\n

product is the sum of the logs. Since logs\n

that when you multiply two numbers together,\n

we said for the exponent version. The quotient\n

of logs by saying the log of x divided by\n

y. In words, we can say that the log of the\n

logs. Since logs are really exponents, another\n

you divide two numbers, you subtract their\n

rule above. Finally, the power rule for exponents\n

the log of x to the n is equal to n times\n

by saying when you take the log of an expression\n

and multiply. If we think of x as being some\n

take a power to a power, we multiply their\n

the power rule above. It doesn't really matter\n

side, or on the right side, but it's more\n

I've given these rules with the base of two,\n

you remember them, please take a moment to\n

you should get the following chart. Let's\n

expressions as a sum or difference of logs.\n

10 of a quotient. So we can rewrite the log\nof the quotient

To keep things simple, we'll write everything\n

rules hold for any base, we know that if we\n

have a product rule for exponents, which says\n

to two to the m plus n. In other words, if\n

We also have a quotient rule that says that\n

to two to the m minus n. In words, that says\n

the exponents. Finally, we have a power rule\n

we multiply the exponents. Each of these exponent\n

first rule, two to the zero equals one can\n

two of one equals zero. That's because log\n

zero equals one. The second rule, the product\n

saying log of x times y equals log of x plus\n

with my base that I'm using for my exponent\n

product is the sum of the logs. Since logs\n

that when you multiply two numbers together,\n

we said for the exponent version. The quotient\n

of logs by saying the log of x divided by\n

y. In words, we can say that the log of the\n

logs. Since logs are really exponents, another\n

you divide two numbers, you subtract their\n

rule above. Finally, the power rule for exponents\n

the log of x to the n is equal to n times\n

by saying when you take the log of an expression\n

and multiply. If we think of x as being some\n

take a power to a power, we multiply their\n

the power rule above. It doesn't really matter\n

side, or on the right side, but it's more\n

I've given these rules with the base of two,\n

you remember them, please take a moment to\n

you should get the following chart. Let's\n

expressions as a sum or difference of logs.\n

10 of a quotient. So we can rewrite the log\nof the quotient

as the difference of the logs. Now we still\n

log of a product as the sum of the logs

as the difference of the logs. Now we still\n

log of a product as the sum of the logs

So that is log of y plus log of z. When I\n

because here I'm subtracting the entire log\n

So that is log of y plus log of z. When I\n

because here I'm subtracting the entire log\n

I'll make sure I do that by putting them in\n

by distributing the negative sign. And here's\n

have the log of a product. So I can rewrite\n

my power rule to bring down the exponent T,\n

the final expression log of five plus

I'll make sure I do that by putting them in\n

by distributing the negative sign. And here's\n

have the log of a product. So I can rewrite\n

my power rule to bring down the exponent T,\n

the final expression log of five plus

t times log of two. One common mistake on\n

as t times log of five times two. In fact,\n

the T only applies to the two, not to the\n

it down in front using the power rule. After\n

expression raised to an exponent, and not\n

we're going to go the other direction. Here,\n

And we want to wrap them up into a single\n

that's a difference of logs. So I know I can\n

have the sum of two logs. So I can rewrite\n

up a little bit and rewrite it as log base\n

I can rewrite the sum of my logs as the log\n

this difference of logs as the log of a quotient,\n

of two multiplied in front. But I can use\n

up in the exponent. So I'll do that first.\n

x minus one, and rewrite this second term\n

I have a straightforward difference of two\n

quotient. I can actually simplify this some\n

the same thing as x squared minus one. I can\n

minus one. In this video, we solve for rules\n

First, we saw that the log with any base of\n

rule, the log of a product is equal to the\n

the log of a quotient is the difference of\n

you take a log of an expression with an exponent\n

multiply it. It's worth noticing that there's\n

of a song. In particular, the log of a psalm\n

think about logs and exponent rules going\n

there's also no rule for rewriting the sum\n

will be super handy. As we start to solve\n

really useful method for finding the derivative\n

start with a brief review of composition.\n

the output of G as a diagram, this means we\n

t times log of two. One common mistake on\n

as t times log of five times two. In fact,\n

the T only applies to the two, not to the\n

it down in front using the power rule. After\n

expression raised to an exponent, and not\n

we're going to go the other direction. Here,\n

And we want to wrap them up into a single\n

that's a difference of logs. So I know I can\n

have the sum of two logs. So I can rewrite\n

up a little bit and rewrite it as log base\n

I can rewrite the sum of my logs as the log\n

this difference of logs as the log of a quotient,\n

of two multiplied in front. But I can use\n

up in the exponent. So I'll do that first.\n

x minus one, and rewrite this second term\n

I have a straightforward difference of two\n

quotient. I can actually simplify this some\n

the same thing as x squared minus one. I can\n

minus one. In this video, we solve for rules\n

First, we saw that the log with any base of\n

rule, the log of a product is equal to the\n

the log of a quotient is the difference of\n

you take a log of an expression with an exponent\n

multiply it. It's worth noticing that there's\n

of a song. In particular, the log of a psalm\n

think about logs and exponent rules going\n

there's also no rule for rewriting the sum\n

will be super handy. As we start to solve\n

really useful method for finding the derivative\n

start with a brief review of composition.\n

the output of G as a diagram, this means we\n

Then we apply f to the output to get our final\n

and F the outer function. Because g looks\n

Then we apply f to the output to get our final\n

and F the outer function. Because g looks\n

in this standard notation, we can write h\n

x as the composition of two functions, by\n

the square root function be the outer function,\n

root of u. I like to do this sort of dissection\n

of the function, whatever is inside the box\n

to the box becomes our outer function, in\n

us to write h of x as the composition, f of\n

functions defined here. Please take a moment\n

of functions, before you go on. A natural\n

let our inner function be tan of x plus seacon\n

to that box the inner function, it gets cubed\n

to write the next example as a composition\n

squared as our inner function, and then our\n

inner function. Alternatively, we could take\n

in this standard notation, we can write h\n

x as the composition of two functions, by\n

the square root function be the outer function,\n

root of u. I like to do this sort of dissection\n

of the function, whatever is inside the box\n

to the box becomes our outer function, in\n

us to write h of x as the composition, f of\n

functions defined here. Please take a moment\n

of functions, before you go on. A natural\n

let our inner function be tan of x plus seacon\n

to that box the inner function, it gets cubed\n

to write the next example as a composition\n

squared as our inner function, and then our\n

inner function. Alternatively, we could take\n

has to be e to the power. It's also possible\n

of three functions. An inner function of x\n

outermost function of e to the power, which\n

has to be e to the power. It's also possible\n

of three functions. An inner function of x\n

outermost function of e to the power, which\n

When calculating the derivatives of complicated\n

them as compositions of simpler functions.\n

terms of the simpler derivatives. And that's\n

rule tells us if we have two differentiable\n

f composed with g of x is equal to the derivative\n

function times the derivative of the inner\n

instead, in lightness notation, that is the\n

let u equal g of x. And let's let y equal\n

When calculating the derivatives of complicated\n

them as compositions of simpler functions.\n

terms of the simpler derivatives. And that's\n

rule tells us if we have two differentiable\n

f composed with g of x is equal to the derivative\n

function times the derivative of the inner\n

instead, in lightness notation, that is the\n

let u equal g of x. And let's let y equal\n

Now do u dx is just another way of writing\n

writing f prime of U. or in other words, f\n

dX, that means we're taking the derivative\n

with g prime of x. Using this key, I can rewrite\n

Now do u dx is just another way of writing\n

writing f prime of U. or in other words, f\n

dX, that means we're taking the derivative\n

with g prime of x. Using this key, I can rewrite\n

times d u dx. These are the two alternative\n

the chain rule to take the derivative of the\n

times d u dx. These are the two alternative\n

the chain rule to take the derivative of the\n

Actually, I'm going to rewrite this as h of\n

it easier to take derivatives. As a composition,\n

x and the outer function as the one half power.\n

Actually, I'm going to rewrite this as h of\n

it easier to take derivatives. As a composition,\n

x and the outer function as the one half power.\n

of x we need to take the derivative of the\n

and then multiply that by the derivative of\n

of the inner function, sine x is just cosine\n

is one half times u to the negative one half.\n

the negative one half. But that's evaluated\n

of x we need to take the derivative of the\n

and then multiply that by the derivative of\n

of the inner function, sine x is just cosine\n

is one half times u to the negative one half.\n

the negative one half. But that's evaluated\n

And then we multiply that by cosine of x.\n

function evaluated on the inner function times\n

we found the derivative using the chain rule.\n

tan x plus seacon X and our outer function,\n

15 times u squared. But that's the evaluated\n

we still need to multiply that by the derivative\n

we get the 15 tan x plus secant x squared\n

squared x, plus the derivative of secant x,\n

rule derivative. In this last example, we're\n

e to the power and its inner function is sine\n

has an outer function of sine and an inner\n

of x, we first have to take the derivative\n

of e to the power is just e to the power.

And then we multiply that by cosine of x.\n

function evaluated on the inner function times\n

we found the derivative using the chain rule.\n

tan x plus seacon X and our outer function,\n

15 times u squared. But that's the evaluated\n

we still need to multiply that by the derivative\n

we get the 15 tan x plus secant x squared\n

squared x, plus the derivative of secant x,\n

rule derivative. In this last example, we're\n

e to the power and its inner function is sine\n

has an outer function of sine and an inner\n

of x, we first have to take the derivative\n

of e to the power is just e to the power.

And now we evaluate that on its inner function,\n

we have to multiply that by the derivative\n

I'll copy down the E to the sine x squared.\n

to find the derivative of sine x squared.\n

of sine is cosine. I need to evaluate it on\n

multiply that by the derivative of the inner\n

have to take the derivative of x squared,\nwhich is 2x

And now we evaluate that on its inner function,\n

we have to multiply that by the derivative\n

I'll copy down the E to the sine x squared.\n

to find the derivative of sine x squared.\n

of sine is cosine. I need to evaluate it on\n

multiply that by the derivative of the inner\n

have to take the derivative of x squared,\nwhich is 2x

This video introduced the chain rule, which\n

g at x is equal to f prime at g of x times\n

to d y d u times d u dx. This video gives\n

the chain rule, and also includes a handy\n

respect to x, where A is any positive number.\n

the chain rule that the derivative of five\n

the x. First, I want to rewrite five to the\n

that because e to the ln five is equal to\n

equal to five to the x. But e to the ln five\n

e to the ln five times x. So if I want to\n

rewriting it as e to the ln five times x,\n

ln five times x. And I'm going to think of\n

That's what I wanted to Make the derivative\n

the derivative of the outer function, derivative\n

and I evaluate it at its inner function. But\n

derivative of the inner function, well, the\n

constant coefficient ln five. And that's my\n

times x is just five to the x. That's what\n

is five to the x times ln five, or I guess\n

is that there's nothing special about five\n

process with any a base positive base a. So\n

that the derivative of a to the x with respect\n

is a fact worth memorizing. I'll use this\n

expression, sine of 5x times the square root\n

dydx, I'll first use the product rule, since\n

expressions. So D y dX is the first expression\n

which I'll go ahead and write using an exponent\n

of the first expression times the second expression.\n

the derivative here. My outermost function\n

the one half power. So when I take the derivative,\n

half, right to the cosine 5x plus one to the\n

I'm going to have to multiply by the derivative\n

cosine 5x plus one, I'll just carry along\n

want to take the derivative of two to the\n

the chain rule again, thinking of my outer\n

So let me copy things down on the next line.\n

of one is just zero, so I'm really just taking\n

by my formula, this is going to be ln f two\ntimes two

This video introduced the chain rule, which\n

g at x is equal to f prime at g of x times\n

to d y d u times d u dx. This video gives\n

the chain rule, and also includes a handy\n

respect to x, where A is any positive number.\n

the chain rule that the derivative of five\n

the x. First, I want to rewrite five to the\n

that because e to the ln five is equal to\n

equal to five to the x. But e to the ln five\n

e to the ln five times x. So if I want to\n

rewriting it as e to the ln five times x,\n

ln five times x. And I'm going to think of\n

That's what I wanted to Make the derivative\n

the derivative of the outer function, derivative\n

and I evaluate it at its inner function. But\n

derivative of the inner function, well, the\n

constant coefficient ln five. And that's my\n

times x is just five to the x. That's what\n

is five to the x times ln five, or I guess\n

is that there's nothing special about five\n

process with any a base positive base a. So\n

that the derivative of a to the x with respect\n

is a fact worth memorizing. I'll use this\n

expression, sine of 5x times the square root\n

dydx, I'll first use the product rule, since\n

expressions. So D y dX is the first expression\n

which I'll go ahead and write using an exponent\n

of the first expression times the second expression.\n

the derivative here. My outermost function\n

the one half power. So when I take the derivative,\n

half, right to the cosine 5x plus one to the\n

I'm going to have to multiply by the derivative\n

cosine 5x plus one, I'll just carry along\n

want to take the derivative of two to the\n

the chain rule again, thinking of my outer\n

So let me copy things down on the next line.\n

of one is just zero, so I'm really just taking\n

by my formula, this is going to be ln f two\ntimes two

to the power of cosine 5x. But of course,\n

to the power of cosine 5x. But of course,\n

and multiply by that the derivative of the\n

I'm just going to carry the rest of the expression\n

taking the derivative of cosine 5x, I think\n

times x is the inner function. Similarly,\n

5x is the inner function. So I can complete\n

now taking the derivative of cosine, which\n

times the derivative of the inner function\n

to that the derivative of sine of 5x. Well,\n

on its inner function times the derivative\n

times the rest of the stuff. I'll do a modest\n

constants out and combine any terms that I\n

example. And the next example, we'll try to\n

value x equals one just based on a table of\n

of f composed with g is just going to be f\n

x, but I want to do this whole process at\n

prime at g of one times g prime of one. Well,\n

at two, and f prime at two is 10. And g prime\n

50. I'm not going to give a rigorous proof\n

a more informal explanation based on the limit\n

write the derivative of f composed with g\n

to a of f composed with g of x minus f composed\n

this slightly. And now we're going to multiply\n

a, that doesn't change the value of expression,\n

That's the detail on sweeping under the rug\n

just a more informal explanation. Now if I\n

the product as the product of the limits,\n

of g. for the limit on the left, notice that\n

since G is differentiable on there for continuous\n

say u be equal to g of x, I can rewrite this\n

minus f of g of A over u minus g of a. Now\n

way of writing the derivative of f evaluated\n

for the chain rule. Let me just emphasize\n

quite airtight, because g of x minus g of\n

more examples of the chain rule justification\n

and multiply by that the derivative of the\n

I'm just going to carry the rest of the expression\n

taking the derivative of cosine 5x, I think\n

times x is the inner function. Similarly,\n

5x is the inner function. So I can complete\n

now taking the derivative of cosine, which\n

times the derivative of the inner function\n

to that the derivative of sine of 5x. Well,\n

on its inner function times the derivative\n

times the rest of the stuff. I'll do a modest\n

constants out and combine any terms that I\n

example. And the next example, we'll try to\n

value x equals one just based on a table of\n

of f composed with g is just going to be f\n

x, but I want to do this whole process at\n

prime at g of one times g prime of one. Well,\n

at two, and f prime at two is 10. And g prime\n

50. I'm not going to give a rigorous proof\n

a more informal explanation based on the limit\n

write the derivative of f composed with g\n

to a of f composed with g of x minus f composed\n

this slightly. And now we're going to multiply\n

a, that doesn't change the value of expression,\n

That's the detail on sweeping under the rug\n

just a more informal explanation. Now if I\n

the product as the product of the limits,\n

of g. for the limit on the left, notice that\n

since G is differentiable on there for continuous\n

say u be equal to g of x, I can rewrite this\n

minus f of g of A over u minus g of a. Now\n

way of writing the derivative of f evaluated\n

for the chain rule. Let me just emphasize\n

quite airtight, because g of x minus g of\n

more examples of the chain rule justification\n

of A to the X is equal to ln of a times a\n

for why the chain rule holds.

of A to the X is equal to ln of a times a\n

for why the chain rule holds.

I'm not going to give a rigorous proof of\n

more informal explanation based on the limit\n

I'm not going to give a rigorous proof of\n

more informal explanation based on the limit\n

So I'm going to write the derivative of f\n

limit as x goes to a of f composed with g\n

by x minus a. I'll rewrite this slightly.\n

the bottom by g of x minus g of a, that doesn't\n

g of x minus g of A is not zero. That's the\n

why this is not a real proof, but just a more\n

and rewrite the limit of the product as the\n

here is just the derivative of g. for the\n

a, g of x has to go to G evey, since G is\n

so I can rewrite this and letting say u be\n

limit as u goes to g of a of f of u minus\nf of g of A

So I'm going to write the derivative of f\n

limit as x goes to a of f composed with g\n

by x minus a. I'll rewrite this slightly.\n

the bottom by g of x minus g of a, that doesn't\n

g of x minus g of A is not zero. That's the\n

why this is not a real proof, but just a more\n

and rewrite the limit of the product as the\n

here is just the derivative of g. for the\n

a, g of x has to go to G evey, since G is\n

so I can rewrite this and letting say u be\n

limit as u goes to g of a of f of u minus\nf of g of A

u minus g of a. Now my expression on the left\n

of f evaluated at G Ave. And I've arrived\n

me just emphasize again, this is just a pseudo\n

x minus g of a might be zero. That's all for\n

complete proof. Please see the textbook implicit\n

the slopes of tangent lines for curves that\n

even functions. So far, we've developed a\n

of functions defined explicitly, in terms\n

section, we'll consider curves that are defined\n

x's and y's. So the points on this curve are\n

As you can see, when you have implicitly defined\n

And in fact, they can not only violate the\n

or be broken up into several pieces or look\n

But small pieces of these curves do satisfy\n

is a function of x. And that allows us to\n

chain rule, to compute derivatives for these\n

u minus g of a. Now my expression on the left\n

of f evaluated at G Ave. And I've arrived\n

me just emphasize again, this is just a pseudo\n

x minus g of a might be zero. That's all for\n

complete proof. Please see the textbook implicit\n

the slopes of tangent lines for curves that\n

even functions. So far, we've developed a\n

of functions defined explicitly, in terms\n

section, we'll consider curves that are defined\n

x's and y's. So the points on this curve are\n

As you can see, when you have implicitly defined\n

And in fact, they can not only violate the\n

or be broken up into several pieces or look\n

But small pieces of these curves do satisfy\n

is a function of x. And that allows us to\n

chain rule, to compute derivatives for these\n

As usual, the derivative dy dx represents\n

example, let's find the equation of the tangent\n

equals 25. drawn below at the point, one,\n

of this tangent line should be about negative\n

So there are at least two ways we could proceed.\n

the same techniques that we've been using.\n

equals 25 minus 9x squared. So y squared is\n

that y is plus or minus the square root of\n

words, plus or minus the square root of 25\n

is giving us the top half of the ellipse.\n

Since the point one, two is on the top part\n

version. And let's take the derivative. But\n

it in a slightly easier form, instead of dividing\n

by the constant one half. And instead of taking\n

an exponent of one half here. So now if I\n

constant of one half. And now I'll start using\n

As usual, the derivative dy dx represents\n

example, let's find the equation of the tangent\n

equals 25. drawn below at the point, one,\n

of this tangent line should be about negative\n

So there are at least two ways we could proceed.\n

the same techniques that we've been using.\n

equals 25 minus 9x squared. So y squared is\n

that y is plus or minus the square root of\n

words, plus or minus the square root of 25\n

is giving us the top half of the ellipse.\n

Since the point one, two is on the top part\n

version. And let's take the derivative. But\n

it in a slightly easier form, instead of dividing\n

by the constant one half. And instead of taking\n

an exponent of one half here. So now if I\n

constant of one half. And now I'll start using\n

taking things to the one half power, and my\n

I'll take the derivative of my outer function\n

inner function to the negative one half. Now\n

function, which is negative 18x. If I simplify\n

over four times 25 minus 9x squared to the\n

negative 9x over two times the square root\n

holds for the top half of the ellipse for\n

negative. Now I want to evaluate the derivative\n

dydx. When x equals one, I get negative nine\n

nine, which is negative nine eighths. Since\n

and I know that point one, two is a point\n

slope form to write down the equation of the\n

negative nine is x plus nine eights plus two,\n

It's now that we've solved the problem once\n

the beginning and solve it again using a new\n

The idea is that I'm going to take the derivative\n

without having to solve for y, I can rewrite\n

of x squared plus four times the derivative\n

of a constant is zero. Going back to the left\n

to x is 2x. Now for the derivative of y squared\n

the chain rule, I'm going to think of taking\n

I'm going to think of y itself as my inside\n

my entire curve is not a function, for small\n

get away with doing this, the derivative of\n

the derivative of my inside function, y as\n

for dy dx, which is going to tell me the slope\n

18x from here, divided by eight y from here,\n

times X over Y. Notice that the formula for\n

it. Of course, for this problem, if I wanted\n

the original equation like I did in method\n

entirely in terms of x, which should be the\n

taking things to the one half power, and my\n

I'll take the derivative of my outer function\n

inner function to the negative one half. Now\n

function, which is negative 18x. If I simplify\n

over four times 25 minus 9x squared to the\n

negative 9x over two times the square root\n

holds for the top half of the ellipse for\n

negative. Now I want to evaluate the derivative\n

dydx. When x equals one, I get negative nine\n

nine, which is negative nine eighths. Since\n

and I know that point one, two is a point\n

slope form to write down the equation of the\n

negative nine is x plus nine eights plus two,\n

It's now that we've solved the problem once\n

the beginning and solve it again using a new\n

The idea is that I'm going to take the derivative\n

without having to solve for y, I can rewrite\n

of x squared plus four times the derivative\n

of a constant is zero. Going back to the left\n

to x is 2x. Now for the derivative of y squared\n

the chain rule, I'm going to think of taking\n

I'm going to think of y itself as my inside\n

my entire curve is not a function, for small\n

get away with doing this, the derivative of\n

the derivative of my inside function, y as\n

for dy dx, which is going to tell me the slope\n

18x from here, divided by eight y from here,\n

times X over Y. Notice that the formula for\n

it. Of course, for this problem, if I wanted\n

the original equation like I did in method\n

entirely in terms of x, which should be the\n

need to do that in order to solve this problem.\n

one and the y value of two to get dy dx at\n

times one half or negative nine, eight, which\n

before. So as before, we can compute the equation\n

y equals negative nine 8x plus 25, eights.\n

was a convenient way to find the derivative.\n

standard methods instead. But in many examples,\n

for y directly. And so implicit differentiation\n

is definitely the key to finding y prime for\n

idea is to take the derivative of both sides\n

pieces. And now use the product rule for the\n

cubed times the derivative of the second function\n

to y, d y dx, don't forget the dydx there,\n

of the first part 3x squared times the second\npart

need to do that in order to solve this problem.\n

one and the y value of two to get dy dx at\n

times one half or negative nine, eight, which\n

before. So as before, we can compute the equation\n

y equals negative nine 8x plus 25, eights.\n

was a convenient way to find the derivative.\n

standard methods instead. But in many examples,\n

for y directly. And so implicit differentiation\n

is definitely the key to finding y prime for\n

idea is to take the derivative of both sides\n

pieces. And now use the product rule for the\n

cubed times the derivative of the second function\n

to y, d y dx, don't forget the dydx there,\n

of the first part 3x squared times the second\npart

Next, I need to take the derivative of sine\n

the derivative of the outside sine is cosine.\n

inside x times y. And that's going to be a\n

plus the derivative of x, which is just one\n

But fortunately, my right hand side is easier.\n

x is 3x squared. And the derivative of y cubed\n

Now I need to solve for the y dx. And since\n

different places, I'm first going to distribute\n

then I'll try to move all the dydx is to the\n

this expression. And now moving alternatives\n

all terms without dydx mm to the right side.\n

I'm going to factor out the dy dx. I'm just\n

finally, I can just divide both sides by all\n

my derivative using implicit differentiation.\n

to find the slopes of tangent lines for curves\n

first to take the derivative of both sides\n

dx. This video is about finding the derivatives\n

that the derivative of the exponential function,\n

what's the derivative of an exponential function\n

one way to find the derivative of an exponential\n

as e to a power. So five, is the same thing\n

log or the log base. See, this makes sense\n

log base e of five, means the power that we\n

e to the ln five, that means we raise e to\n

Well, when you raise E to that power, you\n

as e to the ln five, then that means if we\n

as e to the ln five raised to the x power\n

power to a power, I multiply the exponents.\n

times x. Now I want to take the derivative\n

my rewriting trick, that's the same thing\n

of e to the ln five times x. Now we know how\n

can think of e to the power as our outside\n

function. So now by the chain rule, I take\n

the power, and that's just gives me e to the\n

I stick ln five times x as my inside function,\n

the derivative of the inside function, ln\n

copy over first part, the derivative of a\n

me rewrite this a little bit. So e to the\n

the ln five to the x power, just like before,\n

a power to a power, I multiply the exponent,\n

fancy way of writing five. So I've got five\n

respect to x of five to the x. The same argument\n

function, but for any base exponential function.\n

X for any number A, I'm going to get

Next, I need to take the derivative of sine\n

the derivative of the outside sine is cosine.\n

inside x times y. And that's going to be a\n

plus the derivative of x, which is just one\n

But fortunately, my right hand side is easier.\n

x is 3x squared. And the derivative of y cubed\n

Now I need to solve for the y dx. And since\n

different places, I'm first going to distribute\n

then I'll try to move all the dydx is to the\n

this expression. And now moving alternatives\n

all terms without dydx mm to the right side.\n

I'm going to factor out the dy dx. I'm just\n

finally, I can just divide both sides by all\n

my derivative using implicit differentiation.\n

to find the slopes of tangent lines for curves\n

first to take the derivative of both sides\n

dx. This video is about finding the derivatives\n

that the derivative of the exponential function,\n

what's the derivative of an exponential function\n

one way to find the derivative of an exponential\n

as e to a power. So five, is the same thing\n

log or the log base. See, this makes sense\n

log base e of five, means the power that we\n

e to the ln five, that means we raise e to\n

Well, when you raise E to that power, you\n

as e to the ln five, then that means if we\n

as e to the ln five raised to the x power\n

power to a power, I multiply the exponents.\n

times x. Now I want to take the derivative\n

my rewriting trick, that's the same thing\n

of e to the ln five times x. Now we know how\n

can think of e to the power as our outside\n

function. So now by the chain rule, I take\n

the power, and that's just gives me e to the\n

I stick ln five times x as my inside function,\n

the derivative of the inside function, ln\n

copy over first part, the derivative of a\n

me rewrite this a little bit. So e to the\n

the ln five to the x power, just like before,\n

a power to a power, I multiply the exponent,\n

fancy way of writing five. So I've got five\n

respect to x of five to the x. The same argument\n

function, but for any base exponential function.\n

X for any number A, I'm going to get

a to the x times ln A. Now, you might be wondering,\n

e to the x. So our base here is E. That means\n

sec, ln E, that's log base e of E, that's\n

he? Well, the answer there is one. And so\n

by this new rule we have is e to the x, it\n

attention to the difference between two expressions.\n

x, the variable that we're taking the derivative\n

this exponential function, or we use the derivative\n

eight of the x times ln A. On the other hand,\n

the variable x that we're taking the derivative\n

need this exponential rule. In fact, it doesn't\n

rule, right? dy dx of x cubed would be 3x\n

7x to the sixth and enjoy Add x of x to the\n

the power role. So it's important to pay attention\n

a derivative. In this video, we found that\n

the x is given by ln five times five to the\n

to x of A to the X is going to be ln a times\n

for the derivative of exponential functions.\n

the derivatives of logarithmic functions,\n

log base A x for any positive base a, I want\n

In other words, I want to find the derivative\n

of logarithms, log base a of x equals y means\n

useful because now I can take the derivative\n

Recall of the derivative of a to the power\n

we're thinking of as a function of x, I have\n

rule. The right hand side here is just one.\n

A to the Y. But since age the y is equal to\n

over ln A times x. So the derivative of log\n

of A times x. And in particular, the derivative\n

x. But since ln of E is just one, that saying\n

this is a very handy fact. And this more general\n

we're talking about the derivative of the\n

of the natural log of the absolute value of\n

of x is of course closely related to the function\n

the domain for ln x is just x values greater\n

value of x is all X's not equal to zero. The\n

the graph of y equals ln absolute value of\n

the absolute value of x is equal to x, when\n

x when x is less than zero, ln of the absolute\n

x is greater than or equal to zero, and ln\n

I consider the derivative of ln of absolute\n

of each piece separately. We just saw that\n

derivative of ln of minus x is going to be\n

x, which is minus one by the chain rule. Notice\n

one over x. So the derivative of ln of absolute\n

x is positive or negative. This formula will\n

In this video, we found that the derivative\n

nice derivative. And more generally, the derivative\n

x. We've seen previously that the derivative\n

to the A minus one. This is the power rule.\n

number a raised to the x power is equal to\n

the derivative when the variable x is in the\n

if the variables in both the base and the\n

x to the x. To differentiate functions like\n

differentiation. To find the derivative of\n

to the x. Now we want to find dy dx. Since\n

so let's take the natural log of both sides.\n

you have a variable in the exponent that you\n

of logs allow us to bring that exponent down\n

defined in terms of x, so let's use implicit\n

of both sides with respect to x. And now we\n

because we've gotten rid of the awkward exponential\n

of ln y is one over y times dy dx. And the\n

rule is x times one over x plus one times\nln x.

a to the x times ln A. Now, you might be wondering,\n

e to the x. So our base here is E. That means\n

sec, ln E, that's log base e of E, that's\n

he? Well, the answer there is one. And so\n

by this new rule we have is e to the x, it\n

attention to the difference between two expressions.\n

x, the variable that we're taking the derivative\n

this exponential function, or we use the derivative\n

eight of the x times ln A. On the other hand,\n

the variable x that we're taking the derivative\n

need this exponential rule. In fact, it doesn't\n

rule, right? dy dx of x cubed would be 3x\n

7x to the sixth and enjoy Add x of x to the\n

the power role. So it's important to pay attention\n

a derivative. In this video, we found that\n

the x is given by ln five times five to the\n

to x of A to the X is going to be ln a times\n

for the derivative of exponential functions.\n

the derivatives of logarithmic functions,\n

log base A x for any positive base a, I want\n

In other words, I want to find the derivative\n

of logarithms, log base a of x equals y means\n

useful because now I can take the derivative\n

Recall of the derivative of a to the power\n

we're thinking of as a function of x, I have\n

rule. The right hand side here is just one.\n

A to the Y. But since age the y is equal to\n

over ln A times x. So the derivative of log\n

of A times x. And in particular, the derivative\n

x. But since ln of E is just one, that saying\n

this is a very handy fact. And this more general\n

we're talking about the derivative of the\n

of the natural log of the absolute value of\n

of x is of course closely related to the function\n

the domain for ln x is just x values greater\n

value of x is all X's not equal to zero. The\n

the graph of y equals ln absolute value of\n

the absolute value of x is equal to x, when\n

x when x is less than zero, ln of the absolute\n

x is greater than or equal to zero, and ln\n

I consider the derivative of ln of absolute\n

of each piece separately. We just saw that\n

derivative of ln of minus x is going to be\n

x, which is minus one by the chain rule. Notice\n

one over x. So the derivative of ln of absolute\n

x is positive or negative. This formula will\n

In this video, we found that the derivative\n

nice derivative. And more generally, the derivative\n

x. We've seen previously that the derivative\n

to the A minus one. This is the power rule.\n

number a raised to the x power is equal to\n

the derivative when the variable x is in the\n

if the variables in both the base and the\n

x to the x. To differentiate functions like\n

differentiation. To find the derivative of\n

to the x. Now we want to find dy dx. Since\n

so let's take the natural log of both sides.\n

you have a variable in the exponent that you\n

of logs allow us to bring that exponent down\n

defined in terms of x, so let's use implicit\n

of both sides with respect to x. And now we\n

because we've gotten rid of the awkward exponential\n

of ln y is one over y times dy dx. And the\n

rule is x times one over x plus one times\nln x.

This simplifies to one over y d y dx equals\n

y times one plus ln x. and replacing y with\n

one plus ln x. This technique of taking the\n

for dydx is known as logarithmic differentiation.\n

variables in both the base and the exponent.\n

is in both the base and the exponent. So as\n

that I want to differentiate and compute dydx.\n

my log rules to bring my exponent down and\n

sides with respect to x. On the left, I get\n

one of our x times the derivative of ln tangent\n

of tangent x, or secant squared x, continuing\n

derivative of one of our x, that's going to\n

is minus one times x to the minus

This simplifies to one over y d y dx equals\n

y times one plus ln x. and replacing y with\n

one plus ln x. This technique of taking the\n

for dydx is known as logarithmic differentiation.\n

variables in both the base and the exponent.\n

is in both the base and the exponent. So as\n

that I want to differentiate and compute dydx.\n

my log rules to bring my exponent down and\n

sides with respect to x. On the left, I get\n

one of our x times the derivative of ln tangent\n

of tangent x, or secant squared x, continuing\n

derivative of one of our x, that's going to\n

is minus one times x to the minus

times ln tangent of x. Simplifying the right\n

sine x over cosine x times one over cosine\n

I can flip and multiply to get one over x\n

cosine squared x minus the second term, canceling\n

of cosecant and secant, I get this expression,\n

times ln tangent of x. Simplifying the right\n

sine x over cosine x times one over cosine\n

I can flip and multiply to get one over x\n

cosine squared x minus the second term, canceling\n

of cosecant and secant, I get this expression,\n

So multiplying both sides by y, I get the\n

the one over x, I can rewrite everything in\n

is most useful when taking the derivative\n

the base and the exponent, like in this example,\n

to take the derivative of a complicated product\n

could take the derivative here just by using\n

it's a little easier to take the log of both\n

the log of a product, we get a sum and the\n

and quotients are a lot easier to deal with.\n

to ln of x plus ln of cosine of x minus ln\n

can even bring that fifth power down, because\n

much more straightforward to take the log\n

y dydx, as usual, and on the right, the derivative\n

cosine x is one over cosine of x times negative\n

plus x is one over x squared plus x times\n

times one over x minus sine x over cosine\n

times 2x plus one over x squared plus x. Now\n

be done. Again, I didn't have to use logarithmic\n

I could have just used the product rule in\n

made it computationally much easier. In this\n

of expressions that have a variable both in\n

was first to set y equal to the expression\n

log of both sides. Next, to derive both sides.\n

is called logarithmic differentiation. The\n

does, so the inverse of tying your shoes would\n

that adds two to a number would be the function\n

introduces inverses and their properties.\n

this chart. In other words, f of two is three,\n

f of five is one, the inverse function for\n

what f does. Since f takes two to three, F\n

this f superscript, negative one of three\n

five, F inverse takes five to three. And since\n

And since f takes five to one, f inverse of\n

in the chart. Notice that the chart of values\n

when y equals f inverse of x are closely related.\n

for f of x correspond to the y values for\n

x correspond to the x values for f inverse\n

inverse functions reverse the roles of y and\n

So multiplying both sides by y, I get the\n

the one over x, I can rewrite everything in\n

is most useful when taking the derivative\n

the base and the exponent, like in this example,\n

to take the derivative of a complicated product\n

could take the derivative here just by using\n

it's a little easier to take the log of both\n

the log of a product, we get a sum and the\n

and quotients are a lot easier to deal with.\n

to ln of x plus ln of cosine of x minus ln\n

can even bring that fifth power down, because\n

much more straightforward to take the log\n

y dydx, as usual, and on the right, the derivative\n

cosine x is one over cosine of x times negative\n

plus x is one over x squared plus x times\n

times one over x minus sine x over cosine\n

times 2x plus one over x squared plus x. Now\n

be done. Again, I didn't have to use logarithmic\n

I could have just used the product rule in\n

made it computationally much easier. In this\n

of expressions that have a variable both in\n

was first to set y equal to the expression\n

log of both sides. Next, to derive both sides.\n

is called logarithmic differentiation. The\n

does, so the inverse of tying your shoes would\n

that adds two to a number would be the function\n

introduces inverses and their properties.\n

this chart. In other words, f of two is three,\n

f of five is one, the inverse function for\n

what f does. Since f takes two to three, F\n

this f superscript, negative one of three\n

five, F inverse takes five to three. And since\n

And since f takes five to one, f inverse of\n

in the chart. Notice that the chart of values\n

when y equals f inverse of x are closely related.\n

for f of x correspond to the y values for\n

x correspond to the x values for f inverse\n

inverse functions reverse the roles of y and\n

Next, I'll plot the points for y equals f\n

moment and see what kind of symmetry you observe\n

to the red points, you might have noticed\n

mirror images over the mirror line, y equals\n

of y equals f inverse of x can be obtained\n

over the line y equals x. This makes sense,\n

x. In the same example, let's compute f inverse\n

In other words, we're computing f inverse\n

out. So that's f inverse of three. Since F\n

see is two. Similarly, we can compute f of\n

f of f inverse of three. Since f inverse of\n

F of two, which is three. Please pause the\n

compositions. You should have found that in\n

a number, you get back to the very same number\n

f of f inverse of any number, you get back\n

general, f inverse of f of x is equal to x,\n

This is the mathematical way of saying that\n

look at a different example. Suppose that\n

and guess what the inverse of f should be?\n

does. You might have guessed that f inverse\n

And we can check that this is true by looking\n

root of function, which means the cube root\n

if we compute f inverse of f of x, that's\n

to excellence again. So the cube root function\n

When we compose the two functions, we get\n

be nice to have a more systematic way of finding\n

checking. One method uses the fact that inverses,\n

to find the inverse of the function, f of\n

it as y equals five minus x over 3x. Reverse\n

minus y over three y, and then solve for y.\n

by three y. Bring all terms with wisened to\n

then to the right side, factor out the y and\n

of x as five over 3x plus one. Notice that\n

f inverse are both rational functions, but\n

And then General, f inverse of x is not usually\n

because when we write two to the minus one,\n

minus one of x means the inverse function\n

if all functions have inverse functions, that\n

there always a function that it is its inverse?

Next, I'll plot the points for y equals f\n

moment and see what kind of symmetry you observe\n

to the red points, you might have noticed\n

mirror images over the mirror line, y equals\n

of y equals f inverse of x can be obtained\n

over the line y equals x. This makes sense,\n

x. In the same example, let's compute f inverse\n

In other words, we're computing f inverse\n

out. So that's f inverse of three. Since F\n

see is two. Similarly, we can compute f of\n

f of f inverse of three. Since f inverse of\n

F of two, which is three. Please pause the\n

compositions. You should have found that in\n

a number, you get back to the very same number\n

f of f inverse of any number, you get back\n

general, f inverse of f of x is equal to x,\n

This is the mathematical way of saying that\n

look at a different example. Suppose that\n

and guess what the inverse of f should be?\n

does. You might have guessed that f inverse\n

And we can check that this is true by looking\n

root of function, which means the cube root\n

if we compute f inverse of f of x, that's\n

to excellence again. So the cube root function\n

When we compose the two functions, we get\n

be nice to have a more systematic way of finding\n

checking. One method uses the fact that inverses,\n

to find the inverse of the function, f of\n

it as y equals five minus x over 3x. Reverse\n

minus y over three y, and then solve for y.\n

by three y. Bring all terms with wisened to\n

then to the right side, factor out the y and\n

of x as five over 3x plus one. Notice that\n

f inverse are both rational functions, but\n

And then General, f inverse of x is not usually\n

because when we write two to the minus one,\n

minus one of x means the inverse function\n

if all functions have inverse functions, that\n

there always a function that it is its inverse?

In fact, the answer is no. See, if you can\n

does not have an inverse function. The word\n

is a relationship between x values and y values,\n

there's only one corresponding y value. One\n

inverse function is the function f of x equals\n

function is not a function. Note that for\n

the number negative two, both go to number\n

to send four to both two and negative two.\n

be easier to understand the problem, when\n

Recall that inverse functions reverse the\n

line y equals x. But when I flipped the green\n

red graph. This red graph is not the graph\n

line test. The reason that violates the vertical\n

violates the horizontal line test, and has\n

a function f has an inverse function if and\n

line test, ie every horizontal line intersects\n

for a moment and see which of these four graphs\n

words, which of the four corresponding functions\n

found that graphs A and B, violate the horizontal\n

inverse functions. But graph C and D satisfy\n

represent functions that do have inverses.\n

test are sometimes called One to One functions.\n

if for any two different x values, x one and\n

x two are different numbers. Sometimes, as\n

x one is equal to f of x two, then x one has\n

try to find P inverse of x, where p of x is\n

If we graph P inverse on the same axis as\n

by flipping over the line y equals x. If we\n

can write y equal to a squared of x minus\n

for y by squaring both sides and adding two.\n

plus two, that would look like a parabola,\n

drawn together with another arm on the left\n

function consists only of this right arm,\n

the restriction that x has to be bigger than\n

fact that on the original graph, for the square\n

to zero. Looking more closely at the domain\n

the domain of P is all values of x such that\n

Since we can't take the square root of a negative\n

greater than or equal to two, or an interval\n

The range of P, we can see from the graph\n

zero, or the interval from zero to infinity.

In fact, the answer is no. See, if you can\n

does not have an inverse function. The word\n

is a relationship between x values and y values,\n

there's only one corresponding y value. One\n

inverse function is the function f of x equals\n

function is not a function. Note that for\n

the number negative two, both go to number\n

to send four to both two and negative two.\n

be easier to understand the problem, when\n

Recall that inverse functions reverse the\n

line y equals x. But when I flipped the green\n

red graph. This red graph is not the graph\n

line test. The reason that violates the vertical\n

violates the horizontal line test, and has\n

a function f has an inverse function if and\n

line test, ie every horizontal line intersects\n

for a moment and see which of these four graphs\n

words, which of the four corresponding functions\n

found that graphs A and B, violate the horizontal\n

inverse functions. But graph C and D satisfy\n

represent functions that do have inverses.\n

test are sometimes called One to One functions.\n

if for any two different x values, x one and\n

x two are different numbers. Sometimes, as\n

x one is equal to f of x two, then x one has\n

try to find P inverse of x, where p of x is\n

If we graph P inverse on the same axis as\n

by flipping over the line y equals x. If we\n

can write y equal to a squared of x minus\n

for y by squaring both sides and adding two.\n

plus two, that would look like a parabola,\n

drawn together with another arm on the left\n

function consists only of this right arm,\n

the restriction that x has to be bigger than\n

fact that on the original graph, for the square\n

to zero. Looking more closely at the domain\n

the domain of P is all values of x such that\n

Since we can't take the square root of a negative\n

greater than or equal to two, or an interval\n

The range of P, we can see from the graph\n

zero, or the interval from zero to infinity.

Similarly, based on the graph, we see the\n

or equal to zero, the interval from zero to\n

values greater than or equal to two, or the\n

closely at these domains and ranges, you'll\n

to the range of P inverse, and the range of\n

This makes sense because inverse functions\n

f inverse of x is the x values for F inverse,\n

of F. The range of f inverse is the y values\n

or the domain of f. In this video, we discussed

Similarly, based on the graph, we see the\n

or equal to zero, the interval from zero to\n

values greater than or equal to two, or the\n

closely at these domains and ranges, you'll\n

to the range of P inverse, and the range of\n

This makes sense because inverse functions\n

f inverse of x is the x values for F inverse,\n

of F. The range of f inverse is the y values\n

or the domain of f. In this video, we discussed

five key properties of inverse functions.\n

x. The graph of y equals f inverse of x is\n

the line y equals x. When we compose F with\n

equals x. And similarly, when we compose f\n

words, F and F inverse undo each other. The\n

and only if the graph of y equals f of x satisfies\n

domain of f is the range of f inverse. And\n

These properties of inverse functions will\n

and their inverses logarithmic functions.\n

functions, sine inverse cosine inverse and\n

please focus first on the thin black line.\n

of the inverse of a function can be found\n

over the line y equals x. I've drawn the flipped\n

notice that the blue dotted line is not the\n

vertical line test. So in order to get a function,\n

need to restrict the domain of sine of x,\n

with a thick black line. If I invert that\n

x, I get the piece drawn with a red dotted\n

vertical line test. So it is in fact a function.\n

infinity to infinity, or restricted sine x\n

over two. It's it's range is still from negative\n

Because I've taken the biggest possible piece\n

a function. The inverse sine function is often\n

a function reverses the roles of y at x, it\n

sine of x, the inverse function has domain\n

pi over two to pi over two, which seems plausible\n

the work of a function. So if the function\n

the inverse sine, or arc sine takes numbers\n

of pi over two is one, arc sine of one is\n

arc sine of x is the angle between negative\n

x. y is equal to arc sine x means that x is\n

angles, y who sine is x, right, they all differ\n

y is between negative pi over two and pi over\n

domain restriction in order to get a well\n

notation for inverse sine. Sometimes it's\n

But this notation can be confusing, so be\n

one of x does not equal one over sine of x.

One over sine of x the reciprocal function\n

sign to the negative one of x is another word\n

which is not the same thing as the reciprocal\n

to build an inverse cosine function, we start\n

the line y equals x to get the blue dotted\n

So we go back and restrict the domain for\n

zero and pi. The resulting red graph now satisfies\n

function. Our restricted cosine has domain\n

to one. And so our inverse function, arc cosine\n

from zero to pi. Since cosine takes us from\n

numbers back to angles. For example, cosine\n

over two. So arc cosine of the square root\n

arc cosine of x is the angle between zero\n

y equals r cosine of x means that x is equal\n

pi. Since otherwise, there'd be lots of possible\n

The alternative notation for arc cosine is\n

cosine to the negative 1x is not the same\n

of x is also called secant of x. cosine to\n

function, and these two things are not the\n

tangent function. Here's a graph of tangent\n

part of the function, they're just vertical\n

when we flip over the line y equals x, we\n

Here we've taken the piece marked in black,\n

get this piece in red, which is actually a\n

line test. Now, you might ask, would it be\n

tangent function to invert? And the answer\n

maybe mathematicians do that. But on our planet,\n

of tangent to invert, which is kind of a convenient\n

origin. In the previous two examples, our\n

cosine was also a convention that led to a\n

any case, based on our choice, our restricted\n

two to pi over two. We don't include the endpoints\n

function has vertical asymptotes, that negative\n

there. The range of our restricted tan function\n

arc tan of X has domain from negative infinity\n

two to pi over two. Once again, tangent is\n

is taking us from numbers to angles. For example,\n

arc tan of one is pi over four. So arc tan\n

two and pi over two whose tangent is x.

y is equal to arc tan x means that x is equal\n

pi over two and pi over two. The inverse tan\n

minus one of x. And once again, tan inverse\n

tan of x. And it's not equal to one over tan\n

that's all for this video on the three basic\n

known as arc sine of x, cosine inverse x,\n

x, also known as arc tan x. In this video,\n

the derivatives of the inverse trig functions.\n

y equals sine inverse of x means that y is\n

words, we can write x equals sine of y as\n

equivalent, because there are lots of different\n

can have the same resulting side. And for\n

that angle y has to be between pi over two\n

convention. Be careful not to mistake sine\n

and one over sine x, which is a reciprocal\n

negative one does not mean reciprocal here,\n

notation for inverse sine, which is arc sine.\n

inverse of x, we want to find the derivative\n

derivative of y with respect to x, where y\n

this equation here as x equals sine y, and\n

the derivative of both sides with respect\n

derivative of sine y. In other words, one\n

for dydx, I have that dydx is one over cosine\n

not in a super useful form, because there's\n

it all in terms of x. Well, I could rewrite\n

of x, since after all, y is equal to sine\n

useful form, because it's difficult to evaluate\n

to look at a right triangle. I want to label\n

I'll put it here. And since sine of y is x,\n

label my opposite side with x and my hypothesis\n

length of my remaining side, it's going to\n

squared by the Pythagorean Theorem. Now I\n

Cosine of y is adjacent overhype hotness,\n

over one, or just the square root of one minus\n

y is a positive angle between zero and pi\n

but you can check that the same formula also\n

it going down here on the unit circle instead\nof up here.

So now that I have a formula for cosine y\n

and substitute and I get dy dx is one over\n

other words, I found a formula for With the\n

out a similar process to find the derivative\n

x means that x is equal to cosine of y. And\n

To find the derivative of arc cosine of x,\n

for cosine inverse, I can write y equals arc\n

cosine of y. And then I want to find dydx.\n

the video and try it for yourself before going\n

cosine y, we're going to take the derivative\n

of x is one, and the derivative of cosine\n

equal to negative one over sine y. As before,\n

angle is y. And now I know that x is cosine\n

side and one on my partner's leaving the square\n

side, which means that sine of y, which is\n

square root of one minus x squared. And so\n

square root of one minus x squared. And I\n

cosine of x. inverse tangent can be handled\n

try it for yourself before watching the video.\n

is tangent of y. And the convention is that\n

two and pi over two. Proceeding as before,\n

tan of y, take the derivative of both sides.\n

y dx. Solving for dydx we have one over seacon\n

as before, we can label the angle of as y.\n

is opposite over adjacent, I'm going to label\n

one, which gives us a hypothesis of the square\n

secant of y is one over cosine of y. So that's\n

that's the square root of one plus x squared\n

the square of this, which is one plus x squared.\n

dx, and I get dy dx is one over one plus x\n

the derivative of inverse tangent. The other\n

inverse and cosequin inverse, have derivatives\n

table summarizes these results. In some books,\n

X for the formulas for inverse secant and\n

this makes no difference. And when x is negative,\n

the convention for the range of y for these\n

Notice that the derivatives of the inverse\n

negative signs in front of them and are the\n

functions without the CO that makes it easier\n

formulas. Let's do one example using the formulas\n

of tan inverse of A plus x over a minus x,\n

of tan inverse x. Now, to compute dydx, for\n

outside function is tan inverse, whose derivative\n

we'll need to multiply that by the derivative\n

the first part and take the derivative of\n

rule. So I put the denominator on the bottom\n

hi, the derivative of a plus x with respect\n

of a minus x with respect to x is negative\n

the negative one and the negative sign here\n

I have one plus a plus x squared over a minus\n

canceling in the numerator, I get to a and\n

x squared plus a plus x squared. If I expand\n

two a over two A squared plus 2x squared,\n

is a pretty nice derivative. So now you know\n

And you also know how to find them using implicit\n

When two or more quantities are related by\n

time are also related. That's the idea behind\n

of related rates involving distances. A tornado\n

Phillips Hall at a rate of 40 miles per hour.\n

a speed of 12 miles per hour. How fast is\n

after 15 minutes. In a related rates problem,\n

first, that can help you uncover the geometry\n

related. In this problem, we have a right\n

due east and the bicycles traveling due south\n

the quantities of interest. I'll call the\n

Hall a. Although it starts at 20 miles, it\n

idea to assign it a letter a variable. I'll\n

Hall and the bicycle, a quantity that also\n

the distance between the tornado and the bicycle.\n

distance is changing. In other words, DC dt.\n

relate the quantities of interest. In this\n

that a squared plus b squared equals c squared.\n

between you and the tornado is changing. That's\n

bicycle is traveling and the tornado is moving.\n

work these rates of change into the problem,\n

of this equation with respect to time. That's\n

a squared plus b squared. And that's equal\n

of a B and C as functions of T here, since\n

to a times dA DT using the chain rule

plus two B DB dt. And on the right side, I\n

given to me in the problem to plug in numbers\n

DT since the tornado was moving at a rate\n

the tornado and Philip's Hall is decreasing\n

is negative 40. That negative sign is important\n

is decreasing. Since the bicycle is moving\n

Philips Hall and the bicycle is increasing\n

positive 12. The quantity is a, b and c are\n

t equals 15 minutes or in hours, 0.25 hours,\n

tornado is starts 20 miles away, but it's\n

after a quarter of an hour, it's gone 10 miles,\n

only 10 miles away. And so at the time of\n

is moving at 12 miles per hour. So after a\n

And so at this time, b equals three. Now using\n

plug in a and b and solve for C, we know that\n

squared. So C is going to be the square root\n

equation, we get two times 10 times negative\n

times the square root of 109 times DC dt.\n

72 over two times a squared of 109, which\n

the distance between the tornado and us is\n

These same steps will get you through a variety\n

notes. Don't plug in numbers to send. Any\n

as variables, so you can properly take the\n

be careful to use negative numbers for negative\n

are decreasing, we wouldn't have gotten the\n

40 for the rate of change of the distance\n

rates problem, and found that riding a bicycle\n

In this classic related rates problem, water's\n

to figure out how fast the water is rising.\n

cubic meters per minute, the tank is shaped\n

and a radius of five meters at the top, we're\n

level is rising in the tank. When the water\n

Now let's label some quantities of interest.\n

that stay fixed throughout the problem. Like\n

that are varying with time, I need to use\n

So the height of the water is varying throughout\n

be handy to also talk about the radius of\n

I'll call that our ultimately I want to find\n

So that's DHD T. Next, I want to write down\n

From geometry, I know that the volume of a\n

the height. So the volume of water in the\n

times h since h is the height of the piece\n

squared is the area of that circular base\n

the base even though it's at the top. There's\n

here that comes from similar triangles. From\n

for the little triangle here is the same as\n

other words, we know that our over h is going\n

relationship to eliminate one of the variables\n

which one we want to eliminate. Since we're\n

need to keep the variable h in here. But since\n

you changing, it's a good idea to get rid\n

get r equals five fourths times h, and plug\n

get v equals 1/3 pi times five fourths h squared\n

pi h cubed. Now we're gonna derive both sides\n

get rates of change into the problem. Remember\n

and the height of water as functions of time\n

three h squared DHD t. Now let's plug in numbers\n

T. From our problem, we know that water's\n

meters per minute. So dv dt is three, we're\n

level is rising when the water height is two\n

in those values and solving for a DHD T, we\n

It's pi times three times two squared, which\n

point one five meters per second.

This video solve the related rates problem\n

similar triangles to eliminate one variable.\n

involving rotation and angles. a lighthouse\n

light that makes two revolutions per minute\n

runs north south, and there's a cave directly\n

of light moving along the shore at a point\n

picture. Now let's label it with variables\n

time, the distance between the lighthouse\n

to put a variable for that. But the distance\n

where the light is hitting, that's varying.\n

know how fast the beam of light is moving,\n

x is changing. In other words, we want to\n

of this right triangle made by the beam of\n

angle here between the beam of light and the\n

And the angle up here, I suppose is also changing\n

angle between the East West line and the north\n

90 degrees. Next, we want to write down equations\n

I see a right triangle in a problem, I'm tempted\n

in this case would say one half squared plus\n

problem, it doesn't look like that's going\n

and would relate x and h. But we don't have\n

only rate of change information given to us\n

per minute is indirectly telling us how this\n

beam is making two revolutions per minute,\n

that amounts to a change of four pi radians\n

I'd really like to write down the equation\n

trig, I know that tangent of theta is opposite\n

theta equals x divided by one half. Or in\n

the equation that I need that relates x and\n

with respect to time t. And I get second squared\n

Next, I can plug in numbers and solve for\n

x equals one. We already figured out from\n

dt is four pi. Now secant theta is one over\n

ever had partners, it's reciprocal is high\n

that gives us h over one half. Well, when\n

root of one squared plus a half squared by\n

that by one half. And simplifying, we get\n

one half, which ends up as the square root\n

equation involving derivatives. And we get\n

squared times four pi, four d theta dt

equals two times dx dt. Solving for dx dt,\n

by two, or 10 pi. So what are the units here\n

and our time is in minutes, this is 10 pi\n

to more standard units of miles per hour,\n

by 60 minutes per hour to get 600 pi miles\n

per hour, which is pretty darn fast. In this\n

per minute, to a change in angle per minute.\n

and side length. solving a right triangle\n

and the measures of the angles given partial\n

the length of one side and the measure of\n

right angle is 90 degrees, we need to find\n

A, and the length of the two sides labeled\n

of angle A, let's use the fact that the measures\n

180 degrees. So that means that 49 degrees\n

So A is equal to 180 degrees minus 90 degrees\n

To find the length of the side D we have a\n

fact that tan of 49 degrees, which is opposite\n

tan 49 degrees, which works out to 26.46 units.\n

tan of 41 degrees is 23 over b since now if\n

opposite and B is an adjacent That's a little\n

can write B tan 41 degrees equals 23, which\n

With a calculator that works out again to\n

problem and not say sine or cosine is because\n

side that we're looking for be to the side\n

sine instead, would be saying that sine of\n

which would make it difficult to solve. Next,\n

options, we could use a trig function again,\n

degrees, that's adjacent over hypotenuse,\n

that C is 23 over cosine 49, which works out\n

use the Pythagorean Theorem to find C. Since\n

squared. In other words, that's 23 squared\n

means that C is the square root of that song,\n

the ideas we used were, the sum of the angles\n

tangent of an angle being opposite over adjacent\n

we use the Pythagorean Theorem. This allowed\n

of the triangle, knowing just the side length\n

right angles to begin with. In this next example,\n

the right angle, but we know to have the side\nlengths.

To find the unknown angle theta, we can use\n

hotness, so that's 10 over 15. Cosine is a\n

equation relates our unknown angle to our\n

in our equation to solve for. To solve for\n

10/15, which is 0.8411 radians, or 48.19 degrees.\n

use the fact that sine of fee is 10 over 15\n

a little easier, let's just use the fact that\n

us that fee plus 90 plus 48.19 is equal to\n

we can find x either using a trig function,\n

it using a trig function, we could write down\n

10. To find that using the Tyrion theorem,\n

equals 15 squared. I'll use a Pythagorean\n

root of 15 squared minus 10 squared. That\n

use many of the same ideas as in the previous\n

of the angles is 180. The Pythagorean theorem\n

cosine, we also use the inverse trig functions\n

angle. This video showed how it's possible\n

triangle, and the measures of all the angles\n

measure of one angle and one side or from\n

and facts related to maximum and minimum values.\n

maximum at the x value of C. If f of c is\n

in the domain of f. The point with x&y coordinates\n

And the y value f of c is called the absolute\n

the y value f of c is the highest value that\n

maximum point is just a point where it achieves\n

a function to have more than one absolute\n

for the highest value. But a function has\n

f of x has an absolute minimum that x equals\n

x, for all x in the domain of f. In this case,\n

point. And the y value f of c is called the\n

of x, f of c is now the lowest point that\n

and C SOC, are the coordinates of a point\n

For example, this function has an absolute\n

an absolute minimum point with coordinates\n

here, and just has a domain from zero to four,\n

value of 10 at the absolute maximum point\n

keeps going in this direction, it will not\n

maximum and minimum values can also be called\n

In addition to absolute maximum mins, we can\n

of x has a local maximum at x equals C. If\n

for all x, near C. By near C, we mean there's\n

was true. For our graph of f, we have a local\n

highest point anywhere around since there's\n

point in an open interval around see the point\n

the y value f FC is called a local maximum\n

local minimum at x equals C, if f of c is\n

C. And the point c f of c is called a local\n

a local minimum value. A function might have\n

assuming that the domain is zero to four,\n

Because it's the lowest point anywhere nearby.\n

point. Now turning our attention to local\n

here with coordinates about one two. Since\n

for any x value in an open interval around\n

point of 410 does not count as a local maximum\n

interval on both sides of for the function\n

that sort of technical reason, we end up with\n

maximum point here. local maximum and minimum\n

and minimum values. Please take a look at\n

to mark all local maximum minimum points,\n

points. See if you can find the absolute maximum\n

function. I'm going to mark the local maximum\n

points in red. The function definitely has\na local men

here. Since this is the lowest point anywhere\n

a local max point here. There's also a local\n

a low point and open interval. But that local\n

it half green and half red. There's also a\n

since this point is the as low or lower than\n

is defined in an open interval around three,\n

this point is tied for local minimum, with\n

two and three, there are as low or lower than\n

The point 04 doesn't count as a local max,\n

other side of zero. So there's no open interval\n

absolute maximum because the function gets\n

trend continues, the function f of x has no\n

values just keep getting higher and higher\n

point that I want to consider. And that's\n

Well, it's tempting to say that f has a local\n

point in the ground. But in fact, there is\n

value at three is actually down here, too.\n

point. And if you start looking at points\n

maximums either, because you can always find\n

closer and closer, but don't quite reach this\n

the absolute and local maximum points marked.\n

Well, we just said that there is none. But\n

of this absolute minimum point here. So I'd\n

graph of a function. What do you notice about\n

maximum and minimum points, please pause the\n

maximum minimum points are here, here and\n

f prime of c equals zero. And that the third\n

the function has a corner. A number c is called\n

of c does not exist, or f prime of c exists\n

these local maximum minimum points for this\n

this is true in general, if f has a local\n

number for F. We also say that the point c\n

not to read too much into this statement.\n

or man at C then C must be a critical number.\n

if c is a critical number, then f may or may\n

to keep in mind is the function f of x equals\n

of x is 3x squared, we have that f prime of\n

But notice that F does not have a local maximum\n

In this video, we defined absolute and local,\n

numbers, which are numbers c, where f prime\n

exist. We noted that if f has a local max\n

But not necessarily advice. In this video,\n

second derivative can help us find local maximums\n

that f of x has a local maximum at x equals\n

of x, for all x, in an open interval around\n

C. If f of c is less than or equal to f of\n

In this example, the function f has a local\n

a local minimum at x equals about 10. We've\n

min at x equals C, then f prime of c is equal\n

which f prime of c is zero or does not exist,\n

to be careful, because it is possible for\n

a place where if privacy is equal to zero,\n

or min at x equals C. In fact, this happens\n

f prime of two is zero, but there's no local\n

a moment and try to figure out what's different\n

of x equals two, where there's no local max\n

and 11 where there are local maxes and mins.\n

derivative is positive on the left and positive\n

the derivative is positive on the left and\n

minimum, the derivative is negative on the\n

help motivate the first derivative test for\n

derivative test says that if f is a continuous\n

number, then we can decide if f has a local\n

at the first derivative near x equals C. More\n

is positive for x less than c, and negative\n

something like this. Or maybe like this, your\n

x equals C. If on the other hand, f prime\n

for x greater than c, then our function looks\n

x equals C and so we have local men at x equals\n

both sides of C or negative on both sides\n

point at all at x equals C. Instead, our graph\n

this. The first derivative test is great,\n

just by looking at the first derivative. The\n

for finding local maximum points by using\n

derivative test tells us that if f is continuous\n

to zero, and f double prime of c is greater\nthan zero

then f has a local min at x equals C. If on\n

and f double prime of c is less than zero,\n

that if f double prime of c is equal to zero,\n

test is inconclusive. We might have a local\n

not. So we'd have to use a different method\n

In this video, we introduced the first derivative\n

allow us to determine if a function has a\n

value of x. In this video, I'll work through\n

is, maximum values and minimum values of functions.\n

the absolute maximum and minimum values for\n

from zero to four, these maximum and minimum\n

the interior of the interval, or they could\n

we'll need to check the critical numbers,\n

To find the critical numbers, those are the\n

or does not exist. So let's take the derivative\n

get x squared plus x plus two squared on the\n

high, the derivative of the numerator is one\n

that's 2x plus one. Before we figure out where\n

it a little bit. So we can multiply out the\n

And I'll add together like terms in the numerator,\n

all these steps. So our simplified numerator\n

three. Now that I've simplified the derivative,\n

and where it doesn't exist. Let me clear a\n

of x could not exist, is if the denominator\n

zero and 4x squared plus x plus two is always\n

is never zero on this interval. In fact, it\n

never zero, even if we look outside this interval.\n

the quadratic formula. But in any case, we\n

g prime of x does not exist. So we only have\n

To find where g prime of x is equal to zero,\n

is equal to zero. So I'll set negative x squared\n

both sides there by negative one and a factor\n

one. So these are my critical numbers. But\n

negative one doesn't even lie within my interval,\n

to worry about is x equals 3x equals three\n

an absolute maximum or minimum. So let's figure\n

x that evaluates to to 14th, or one set So\n

go ahead and check the endpoints. Those are\n

since our interval is from zero to four. Plugging\n

half, and g of four is 320 seconds. I sometimes\n

The candidate x values are 03, and four and\n

negative a half 1/7, and 320 seconds. Now\n

all I have to do is figure out which one of\n

the smallest. Well, clearly negative one half\n

value. And we just need to compare 1/7 and\n

is the same as 320 firsts, which is going\n

1/7 is our absolute max value. We can confirm\n

g. Remember, we're just interested in the\n

interested in this section of the graph. And\n

at when x equals zero minimum value of negative\n

Well, I'm not sure exactly where it is from\n

around three. And that is a value of something\n

we found as a more precise answer using calculus.\n

extreme values for the function f of x, which\n

on the interval from negative two to two.\n

by checking first the critical numbers, and\n

two and two. To find the critical numbers,\n

But because our function involves the absolute\n

Instead, let's first rewrite f using piecewise\n

the absolute value of x, when x is bigger\n

is just x. So f of x will be x minus x squared.\n

the absolute value of x is negative x. So\n

Now to take the derivative, we can take the\n

than zero, I don't want to take the derivative\n

funny things happening, you know, a cost per\n

x is bigger than zero, and when x is less\n

zero, I can just use the power rule I get\n

than zero, I get negative one minus 2x. And\n

I need to find where f prime of x is equal\n

f prime of x equals zero, where one minus\n

where one, negative one minus 2x is equal\n

to x equal one half for x bigger than zero,\n

than zero. So those are my first two critical\n

interval that I'm interested in. But I also\n

not exist. And the candidate x value for that\n

the derivative does not exist when x equals\n

is very close to one for x values, very close\n

to negative one for x values from the left\n

to have to be sloping down with a slope near\n

a slope near one for x greater than zero,\n

Also notice that even if I weren't 100% sure\n

zero, it's not going to hurt to consider this\n

for the absolute max or min value. So I've\n

points are just going to be x equals negative\n

of values my x values to consider are negative\n

two, and my corresponding f of x values are\n

two minus negative two squared works out to\n

in negative one half, I get one half minus\n

and plugging in one half, I get 1/4. And plugging\n

biggest value is going to be 1/4. So that's\n

is going to be negative two. So that's my\n

looking at the graph. So here I've graphed\n

minus x squared on the interval from negative\n

absolute min is going to be a value of negative\n

and my absolute maximum is going to be a value\n

maximum points. And that concludes this video\n

theorem relates the average rate of change\n

rate of change, or derivative. Let's assume\n

a, b, and maybe defined in some other places\n

whole closed interval. And that is differentiable\n

mean value theorem says that there must be\n

the average rate of change of f on a B is\n

we can write the average rate of change as\n

has to equal f prime at C for some number\n

of f is the slope of the secant line. And\n

some number c, somewhere in between a and\n

exactly the same as the slope of the tangent\n

not necessarily unique. So I encourage you\n

a graph of a function where there's more than\n

drawn something maybe like this. Now, if we\n

c, where the slope of the tangent line is\n

this example, we're asked to verify the mean\n

a particular interval. Verify means that we\n

hold. And also the the conclusion holds. The\n

closed interval, one three, and that is differentiable\n

these facts are true, because f is a polynomial.\n

of the mean value theorem holds. In other\n

one, three, such that the derivative of f\n

of f on the interval from one to three.

Now f prime of x is 6x squared minus eight.\n

minus eight. We can also compute f of three\n

is negative five. Plugging in these values\n

minus eight has to equal 31 minus negative\n

minus eight had better equal 18, which means\n

squared has to equal four, which means that\n

two is not in the interval from one to three,\n

So C equals two is the number we're looking\n

18, which is the average rate of change of\n

value theorem. In this example, we're told\n

of f is bounded between negative three and\n

asked to find the biggest and smallest values\n

mean value theorem gives us one way of relating\n

on the endpoints of the interval. More specifically,\n

rate of change F of six minus f of one over\n

prime of c, for some C, in the interval, one,\n

negative three and negative two, we know that\n

negative three and negative two. We know that\n

inequality for f of six. Multiply the inequality\n

that negative eight is the smallest possible\n

largest possible roles, there is an important\n

f is a function defined on the closed interval\n

closed interval, differentiable on the interior\n

then there's a number c in the interval a,\n

at a graph of such a function that has equal\n

derivative has to be zero at a maximum, or\n

rolls there is a special case of the mean\n

theorem would say about this function, it\n

of c is equal to the average rate of change\n

A are the same, by our assumption, this average\n

value theorem, its conclusion is that there's\n

is exactly the conclusion of rules theorem.\n

that's continuous on a closed interval, and\n

the average rate of change of the function\n

of the function, f prime of c for some C in\n

of the mean value theorem for integrals. the\n

for continuous function f of x, defined on\n

c between A and B, such that f of c is equal\n

that I'm going to give uses the intermediate\n

value theorem says that if we have a continuous\n

call x 1x, two, if we have some number l in\n

has to achieve the value l somewhere between\n

value theorem, let's turn our attention back\n

it's possible that our function f of x might\n

if that's true, then our mean value theorem\n

just equal to that constant, which is equal\n

assume that f is not constant, will it continue\n

to have a minimum value and a maximum value,\n

we know that F's average value on the interval\n

minimum value. If you don't believe this,\n

the interval have to lie between big M and\n

we get little m times b minus a is less than\n

or equal to big M times b minus a. Notice\n

just integrating a constant. Now if I divide\n

little m is less than or equal to the average\n

as I wanted. Now, I just need to apply the\n

as my number L and little m and big M as my\n

value theorem says that F average is achieved\n

x two. And therefore, for some C in my interval\n

for integrals. Now I'm going to give a second\n

And this time, it's going to be as a corollary\n

Recall that the mean value theorem for functions,\n

and differentiable on the interior of that\n

interval, such that the derivative of g at\n

G, across the whole interval from a to b.\n

in mind, and turn our attention back to the\n

to define a function g of x to be the integral\n

given to us in the statement of the mean value\n

is just the integral from a to a, which is\n

to b of our function. Now, by the fundamental\n

continuous and differentiable on the interval\n

And by the mean value theorem for functions,\n

b minus g of a over b minus a, for some numbers,\n

the three facts above, into our equation below,\n

a to b of f of t dt minus zero over b minus\n

wanted to reach. This shows that the mean\n

mean value theorem for functions where our\n

the second proof of the main value theorem\n

value theorem for integrals in two different\n

of calculus along the way in this video, We'll\n

this one, and inequalities involving rational\n

a simple example, maybe a deceptively simple\n

is less than four, you might be very tempted\n

get something like x is less than two as your\n

see why it's not correct, consider the x value\n

inequality, x is less than two, since negative\n

the inequality x squared is less than four,\n

not less than four. So these two inequalities\n

a quadratic inequality just to take the square\n

part of why this reasoning is wrong, as we've\n

we had the equation, x squared equals four,\n

x equals negative two would be another solution.\n

should take this into account. In fact, a\n

x squares or higher power terms, is to solve\n

we even do that, I like to pull everything\n

zero on the other side. So for our equation,\n

x squared minus four is less than zero. Now,\n

equation, x squared minus four is equal to\n

two times x plus two is equal to zero. And\n

x equals two and x equals minus two. Now,\n

on the number line. So I write down negative\n

expression x squared minus four is equal to\nzero.

Since I want to find where x squared minus\n

this expression x squared minus four is positive\n

to plug in test values. So first, I plug in\n

something less than negative two, say x equals\n

into x squared minus four, I get negative\n

four, which is five, that's a positive number.\n

minus four is positive. And in fact, everywhere\n

is going to be positive, because it can jump\n

a place where it's zero, I can figure out\n

negative on this region, and on this region\n

similar way, evaluate the plug in between\n

00 squared minus four, that's negative four\n

x squared minus four is negative on this whole\n

like x equals 10, something bigger than two,\n

computing that I can tell that that's going\n

important. Again, since I want x squared minus\n

the places on this number line where I'm getting\n

line. It's in here, not including the endpoints,\n

x squared minus four is equal to zero and\n

my answer. As an inequality, negative two\n

notation as soft bracket negative two, two\n

similarly, first, we'll move everything to\n

minus 5x squared minus 6x is greater than\n

equation by factoring. So first, I'll write\n

x. And now I'll factor the quadratic. So the\n

six and x e equals negative one, I'll write\n

line. So that's negative one, zero, and six.\n

six times x plus one is equal to zero. But\n

equal to zero. So again, I can use test values,\n

two, either to this version of expression,\n

care whether my answer is positive or negative,\n

version. For example, when x is negative two,\n

x minus six is also negative when I plug in\n

I plug in negative two for x, that's negative\n

times a negative times a negative gives me\n

between negative one and zero, say x equals\n

negative for this factor, a negative for this\n

Negative times negative times positive gives\n

and six, let's try x equals one. Now I'll\n

for this factor, and a positive for this factor.\n

gives me a negative. Finally, for a test value\n

that's going to give me positive positive\n

Since I want values where my expression is\n

places where n equals zero. And the places\n

be close bracket negative one to zero, close\n

As our final example, let's consider the rational\n

by x minus one is less than or equal to zero.\n

denominator and multiply both sides by x minus\n

minus one could be a positive number. But\n

you multiply both sides by a negative number,\n

it's possible to solve the inequality this\n

is less than zero or bigger than zero, I think\n

as we did before. So we'll start by rewriting\n

have zero on the right, well, that's already\n

associated equation. That is x squared plus\n

zero. That would be where the numerators 0x\n

so we're x plus three squared is zero, or\n

step we have to do for rational expressions.\n

does not exist. That is, let's find where\n

equals one. I'll put all those numbers on\n

expression is equal to zero, and the place\n

then I can start in with test values. For\n

work. If I plug those values into this expression\n

answer and a positive answer. The reason I\n

where my denominator is zero is because I\n

to positive by passing through a place where\n

as passing by passing flew to a place where\n

I'm looking for where my original expression\n

I want the places on the number line where\n

places where it's negative. So My final answer\n

negative infinity to one. In this video, we\n

by making a number line. And using test values\n

derivative of a function can tell us a lot\n

In this video, we'll see what f prime and\n

function is increasing and decreasing, is\n

points. We say that a function is increasing.\n

x one is less than x two. In other words,\n

from left to right, we say the function f\n

f of f two, whenever x one is less than x\n

goes down as we move from left to right. In\n

happening when x is near two, is it completely\n

If we assume it's slightly increasing, then,\n

ranges from zero to six, and again as x ranges\n

x values between six and 10. And for x values\n

f can tell us where the function is increasing\n

of x is greater than zero for all x on an\n

This makes sense, because f prime being greater\n

slope. Similarly, if f prime of x is less\n

f is decreasing on this interval. That's because\n

A precise proof of these facts can be found\n

that a function is concave up on an interval\n

a bowl that could hold water on that interval.\n

on that interval. If all the tangent lines\n

the graph of the function. The function is\n

If informally, it looks like an upside down\n

Or more formally, the function is concave\n

graph of the function on an interval. In this\n

again around here. On the left piece, it looks\n

So we can say that f is concave up on the\n

from eight to 11. f is concave down on this\n

can say that f is concave down on the interval\n

from 11 to 12. The concavity of a function\n

where the function is concave up, its derivative\n

values. So the first road was increasing,\n

On this section of the graph, which is also\n

values to zero. That's an increase in the\n

derivative here must be positive. And in this\n

zero to positive values, the first derivative\n

is also positive. On the parts of the function\n

this example, the second derivative is negative.\n

towards zero, that's a decrease in the first\n

Here the first derivative is going from positive\n

first derivative or a negative second derivative.\n

here. In general, we can use the second derivative\n

concavity test says that if the second derivative\n

the function f is concave up on that interval.\n

for all x on the interval, then the function\n

to remember the concavity test is that a positive\n

the smile is supposed to be a concave up function.\n

a sad face where the smile or the frown, I\n

talk about inflection points. A function has\n

continuous at C, and it changes concavity\n

point at x equals C. If f changes from concave\n

from concave down to concave up. In this graph,\n

see that F has an inflection point at x equals\n

down to concave up at x equals four, where\n

down at x equals eight, and again, at x equals\n

derivative being positive or negative, inflection\n

changes sign from positive to negative, or\n

what the inflection point test says. If f\n

at x equals C, then f has an inflection point\n

positive to negative or negative positive,\n

go through a point where it doesn't exist.\n

double prime is zero, it doesn't exist, does\n

inflection point, because it could be zero\n

on both sides. For example, if f of x is x\n

cubed, and f double prime of x is 12x squared.\n

But there is no inflection point at x equals\n

x to the fourth looks kinda like a flattened\n

cavity, f is concave up on both sides of x\n

derivative can tell us where the function\n

derivative can tell us where the function\n

derivative, changing sign from positive to\n

us where we have inflection points. Since\n

complicated functions, it can be extremely\n

value with its tangent line. That's the central\n

Suppose that F of T is the temperature in\n

where t equals zero represents midnight. Suppose\n

f prime of six is three degrees per hour.\n

at 7am and at 8am? Please pause the video\n

at 6am is 60 degrees. So the temperature at\n

60 degrees. But we can do better than this.\n

it's rising at a rate of three degrees per\n

by 7am, the temperature will have risen three\n

the temperature will have had two hours to\n

three degrees per hour. So f of eight should\n

per hour times two hours or 66 degrees. These\n

both the value of the temperature at six,\n

estimates mean graphically, in terms of the\n

over time. And I'll also draw in the tangent\n

of the function and the tangent line is equal\n

three degrees per hour. So that's a rise of\n

which is one hour, after six o'clock, the\n

at eight o'clock, the tangent line has risen\n

our tangent line has had 63 degrees, and at\n

66 degrees. When making these estimates, here,\n

approximate our function. Our actual temperature\n

tangent line, or it possibly could be rising\n

way, the tangent line is a good approximation\n

The idea of approximating a function with\n

that works for any differentiable function.\n

and let A be an arbitrary x value. Let's suppose\n

F of A. And let's say we want to find the\n

it a plus delta x where delta x means a small\n

x directly, we can try to approximate it using\nthe tangent line.

We know that the tangent line has a slope\n

by a run of delta x, the tangent line goes\n

So the height of the tangent line is going\n

x. The linear approximation principle says\n

our tangent line. In other words, f of a plus\n

f prime of a delta x. Remember that delta\n

if you get too far away from a, your tangent\n

of your function. But how small is small enough\n

approximation principle is written with different\n

so x is a number close to a, then delta x\n

principle, as f of x is approximately f of\n

on the right side here is sometimes referred\n

of f at A. That is the linearization of f\n

f prime of A times x minus a. So the approximation\n

approximately equal to l of x. Let's look\n

equation and what it has to do with the tangent\n

down the equation of the tangent line at x\n

can be given in point slope form as y minus\n

naught. Since we're looking for the tangent\n

we can set x naught equal to a and y naught\n

line is just f prime of a. So we can rewrite\n

times x minus a solving for y, we get y equals\n

So this equation for the tangent line is really\n

But linearization is really just a fancy word\n

and definitions on this page. But there's\n

to remember. And that's the idea that you\n

line. If you can keep that idea and this picture\n

approximation principle. And its alternative\n

in an example, the approximation principle\n

equal to f of a plus f prime of A times delta\n

what a should be, and what delta x should\n

root of 59, it makes sense to make our function\n

to pick something that is easy to compute\n

that is easy to compute the square root of\n

64. Since we're trying to compute the square\n

In other words, 64 plus delta x is 59. And\n

to have a negative number for delta x. Now\n

have f of 59 is approximately equal to f of\n

Since f of x is the square root of x, or in\n

of x is going to be one half x to the minus\n

square root of x. So f prime of 64 is one\n

is 1/16. I can rewrite my red equation to\n

the square root of 64 plus 1/16 times negative\n

or 7.6875. using a calculator, I can get a\n

says 7.68114575, up to eight decimal places.\n

this approximation. We have the square root\n

at the tangent line. Our delta x here is a\n

we're using the value of our tangent line\n

root function right here. As you can see from\n

value should be slightly bigger than the actual\n

next example is very similar. We call it the\n

for its tangent line. Namely, the linearization\n

a. And the approximation principle says that\n

to its linearization. Its tangent line, at\n

the same formula that we use in the last problem,\n

of a plus delta x. Since we're trying to estimate\n

function be sine of x. For a, we want to pick\n

it's easy to calculate sine of that number.\n

So let's make a equal to 30 degrees, but let's\n

in calculus, we pretty much always want to\n

when taking derivatives. Since the derivative\n

works when x is in radians, our x needs to\n

want to estimate the sine of, we need to multiply\n

So that becomes 11 pi over 60 radians. Let's\n

And then we'll plug in for x next. So the\n

be sine of pi over six, plus the derivative\n

six. That is l of x is one half, since sine\n

pi over six times x minus pi over six, cosine\n

over two. So this is our equation for the\n

of x at pi over six. Now we know that sine\n

as long as x is near pi over six. So in particular,\n

11 pi over 60 is approximately equal to one\n

times 11 pi over 60 minus pi over six.

That simplifies to one half plus the square\n

now I'm going to cheat a little bit and use\n

of about 0.5453. Now if I use my calculator\n

that's the same thing as sine of 33 degrees.\n

So you can see our approximation using the\n

more accurate value. Notice that in this example,\n

is slightly higher than the actual value.\n

The tangent line at pi over six lies slightly\n

approximate value based on the linearization\n

of sine of 33 degrees. In this video, we use\n

The main formulas were the approximation principle,\n

The key idea is that a differentiable function\n

A by the tangent line at x equals a. The differential\n

familiar concept of approximating a function\n

look familiar from the previous video on linear\n

we have a differentiable function, f of x,\n

a. That is, we know the value of f evey, but\n

x value a plus delta x. That is we don't know\n

line to f of x at x equals a. And we use the\n

for the function at a plus delta x. Since\n

the rise of a run is f prime of a. So if this\n

prime of A times delta x. So the height of\n

going to be f of a plus f prime of A times\n

the extra height here. And since we're using\n

we say that f of a plus delta x is approximately\n

x equivalently. If I subtract F of A from\n

f of a is approximately equal to f prime of\n

approximation principle that we've seen before.\n

So there's nothing new yet. But now I'm going\n

concept. The differential dx is another way\n

a small change in the value of x. The differential\n

f prime of x delta x. Sometimes this is written\n

thing here as df. Sometimes it's handy to\n

value of x, like a value of x equals a, and\n

or f prime of a delta x. Notice that the value\n

Finally, the change in f, which is written\n

minus f of x for some value of x, for example,\n

also be written as the change in y. Using\n

our approximation principle to say, delta\n

in the function is approximately equal to\n

written as the change in Y is approximately\n

d x for the run, and D, F, for the rise of\n

a moment to find delta f in this picture,\n

f of a, so that's this height here. I'll write\n

principle, written in differential notation,\n

delta f is approximated by the rise in the\n

and an example. For the function f of x equal\n

df. We know that df is equal to f prime of\n

is equal to x times the derivative of ln x,\n

x, which is one times ln x. So in other words,\n

plus ln x times dx. When x equals two, and\n

delta x is the same thing as dx, we can just\n

ln of two times negative 0.3. as a decimal,\n

delta f is defined as f of x plus delta x\n

x plus delta x times ln of x plus delta x\n

for x and delta x, we get delta f is two minus\n

two, which according to my calculator is negative\n

in the function between two and two minus\n

change in the tangent line. As expected. The\n

as in this example, suppose that the radius\n

with a possible error of point five centimeters.\n

like this, but the actual sphere might be\n

to use the differential to estimate the resulting\n

Well, the volume of a sphere is given by four\n

If our radius changes by point five centimeters,\n

And that change in volume is the error the\n

of computing delta V directly, we're asked\n

So we're going to use the fact that delta\n

By definition, dv is equal to the derivative\n

as a function of r times Dr. Now, v prime\n

taking the derivative. And here, we're interested\n

Same as delta r of 0.5 centimeters. So dv\n

I plug in R and D R, I get four pi times eight\n

a decimal 402.1 centimeters. That's our error\n

our original error of point five in measuring\n

is its error over the original value of the\n

in volume over the actual volume. Since we're\n

the relative error as dv over V. Now, the\n

thirds times pi times eight cubed. And dv,\n

times 0.5. So dv divided by V is given by\n

an 18.75% relative there. To me, the relative\n

the absolute error estimate above. This video\n

said that we could think of dx as just being\n

the rise in the tangent line, and is equal\n

is the rise in the actual function F. And\n

the picture, dx is the run, df is the rise\n

in the actual function. In the language of\n

principle to say that the change in f can\n

past, we've encountered limits, like the limit\n

minus four. We can't evaluate this limit just\n

two goes to zero, and x squared minus four\n

as a zero over zero indeterminate form. It's\n

what the limit is going to be just by the\n

denominator goes to zero. It depends on how\n

going to zero compared to each other. And\n

number at all, or it could be infinity or\n

been able to evaluate some limits in zero\n

tricks to rewrite the quotients. In this video,\n

a very powerful technique for evaluating limits\n

form the limit as x goes to a of f of x over\n

form, if the limit as x goes to a of f of\n

to a of g of x is equal to zero. A limit and\n

and determinant form. If the limit as x goes\n

infinity. And the limit as x goes to a of\n

We saw an example of a zero over zero indeterminate\n

One example of a infinity over infinity and\n

infinity of 3x squared minus 2x plus seven\n

that as x goes to infinity, the numerator\n

to negative infinity. In these definitions\n

to be negative infinity or infinity, like\n

to be loopy. talls rule can be applied when\n

the derivative of g is nonzero in some open\n

these conditions, if the limit as x goes to\n

or infinity over infinity indeterminant form\n

g of x is the same thing as the limit as x\n

x, provided that the second limit exists,\n

loopy tiles rule in action. In this example,\n

to infinity and the denominator three to the\n

over infinity indeterminate form. So let's\n

limit should equal the limit as x goes to\n

which is one divided by the derivative of\n

three to the x, provided that the second limit\n

In the second limit, the numerators just fixed\n

as x goes to infinity. Therefore, the second\n

evaluates to zero as well. In this example,\n

because as x goes to zero, sine of x and x,\n

of x cubed goes to zero in the denominator.\n

instead, the limit I get by taking the derivative\n

denominator, the derivative of sine x minus\n

of sine x cubed is three times sine x squared\n

me try to evaluate the limit again, as x goes\n

here goes to zero. As x goes to zero, sine\n

one, so the denominator also goes to zero.\n

form. And I might as well try applying loopy\n

point out that cosine of x is going to one.\n

my limit. And in fact, I could rewrite my\n

where the second limit is just one and can\n

rule on this first limit, which is a little\n

the derivative of the top is minus sine x.\n

sine x times cosine x. Now let's try to evaluate\n

going to zero, and our denominator is also\n

to apply lobby towels rule again, because\n

the sine x on the top cancels with the sine\n

limit as the limit of negative one over six\n

In this example, I want to emphasize that\n

of lopi talls rule. If you don't simplify,\n

to apply loopy towels rule and additional\n

the problem more complicated. Instead of simpler\n

zero over zero and infinity over infinity\n

of f of x over g of x with the limit of f\n

second limit exists. This trick is known as\n

can be used to evaluate limits of the form\n

In this video, we'll continue to use lopi\n

forms, like zero times infinity, infinity\n

In this example, we want to evaluate the limit\n

from the positive side, sine x goes to zero,\n

the graph of y equals ln x. So this is actually\n

Even though the second factor is going to\n

times infinity and indeterminate form, you\n

for either positive or negative infinity.\n

zero, the sine x factor is pulling the product\n

the product towards large negative numbers.\n

the product will actually be. But the great\n

to look like an infinity over infinity and\n

determinant form. Instead of sine x times\n

by one over sine x. Now as x goes to zero,\n

And since sine x is going to zero through\n

sine x is going to positive infinity. So I\n

form. Now, I could instead choose to leave\n

put a one over ln x in the denominator. If\n

positive numbers, sine x goes to zero. And\n

over ln x goes to zero. And so I have a zero\n

can be difficult to decide which of these\n

One rule of thumb is to take the version that\n

the numerator and denominator. Another trick\n

get stuck, go back and try the other. I'm\n

it because I recognize that one over sine\n

how to take the derivative of cosecant x.\n

infinity and determinant form, I can rewrite\n

take the derivative of the numerator, that's\n

denominator, that's negative cosecant x cotangent\n

before going any further. I can rewrite my\n

of sine and cosine. cosecant x is one over\n

of x. Now flipping and multiplying, I get\n

x times sine squared of x over negative cosine\n

sine squared x over x cosine x

we know that cosine of x goes to one as x\n

limit of negative sine squared x over x times\n

I once again have a zero over zero indeterminate\n

taking the derivative of the top, I get negative\n

bottom is just one. Now I'm in a good position\n

zero for x in the numerator, I get negative\n

is just one, so my final limit is zero. In\n

x is going to infinity, one over x is going\n

one, but the exponent x is going to infinity,\n

If we had one, to any finite number, that\n

than one, as we raise it to a bigger and bigger\n

our limit has an independent permanent form,\n

to be one infinity, or maybe something in\n

a variable in the base, and a variable in\n

If we set y equal to one plus one over x to\n

sides, I can use my log roles to rewrite that\n

I wanted to take the limit as x goes to infinity\n

x times ln one plus one over x. As x goes\n

One plus one over x goes to just one and ln\n

times zero indeterminate form, which we can\n

or a zero over zero indeterminate form. Let's\n

over x divided by one over x. This is indeed\n

can use lobi tiles rule and take the derivative\n

of the top is one over one plus one over x\n

would be negative one over x squared. And\n

of one over x is negative one over x squared,\n

and rewrite our limit as the limit as x goes\n

x, which is just equal to one, since one over\n

of ln y is equal to one, but we're really\n

we can think of as e to the ln y. Since ln\n

to e to the one. In other words, E. So we\n

E. And in fact, you may recognize that this\n

In the previous example, we had a one to the\n

took logs and use log roles to write that\n

form. Well, the same thing can be done if\n

a zero to the zero indeterminate form. So\n

and zero to the zero, are all indeterminate\n

rule. In this video, we saw that a zero times\n

to a zero over zero, or infinity over infinity\n

g of x as f of x divided by one over g of\n

We also saw how to use lopi talls rule on\n

taking the ln of y, where y is our f of x\n

of. This video is about Newton's method for\n

other words, the values of x that make f of\n

also be thought of as the x intercepts of\n

to the equation either the x equals 4x. This\n

methods. For example, taking the ln of both\n

get x equals ln of 4x, which is just as hard\n

solutions. Looking at the graph of y equals\n

be two solutions, one at approximately x equals\n

x equals maybe point three or point four.\n

more accurate approximations to the solution\n

at the graph. To use Newton's method, instead\n

4x. We'll look at the equation E to the X\n

the function f of x to be e to the x minus\n

all, finding a zero of this function is the\n

equation. So now we're trying to solve the\n

the function f of x equals e to the x minus\n

I'm going to focus on this zero, the one near\n

guess anything reasonably close to the actual\n

guess right here, and I'll call it x one.\n

and I'll write the point on the graph above\n

for the zero of my function, I'm going to\n

that goes through this point. So the second\n

the tangent line is a reasonably good approximation\n

line crosses the x axis should be closer to\n

the x axis, which is the point I'm looking\n

x intercept for the tangent line. I'll call\n

to repeat this process. I'll use x two as\n

where I have the point x to f of x two, and\n

a new intercept. I can repeat this process\n

accurate approximation to my actual zero of\n

graphically, let's find some equations that\n

the initial guess, of x one, then the tangent\n

the x Bayesian y equals f of x one plus f\n

You might remember this equation from a section\n

from the formula y minus y one equals m x\n

m here is the derivative at x one, and y one\n

we have y minus f of x one equals f prime\n

to y equals f of x one plus f prime of x one\n

linearization equation comes from. It's just\n

we want to find the x intercept of the tangent\nline

the tangent line equation equal to zeros,\n

of x one times x minus x one, and we solve\n

each side, divided by f prime of x one and\n

x two. So x two is x one minus f of x one\n

guess, x two, and we can again find the tangent\n

line will be given by the same sort of equation.\n

algebraic steps, get us to the analogous equation\n

f prime of x two. And more generally, as we\n

n plus one guess is going to be given by x\n

f prime of x n. That's the J equation at the\n

the theory down, let's grind through the problem\n

the equation f of x equals e to the x minus\n

So from Newton's methods equation, we have\n

e to the x sub n minus four times X sub n\n

with for example, x sub one equals three,\n

e cubed minus four times three over e cubed\n

I get x sub two equals 2.4973, and so on.\n

take this whole number and plug that in to\n

But for accuracy, when I actually computed\n

my calculator gives me this answer for x of\n

get x of 4x 5x sub six. If I compute one more,\n

to my value in the number of digits that the\n

Newton's methods iterations have converged.\n

eight decimal places. I found one zero for\n

zero, the one over here, I would just need\n

to this x coordinate, perhaps an initial value\n

an algorithm for getting increasingly accurate\n

The central equation that we used was this\n

X sub n to the next 1x sub n plus one. When\n

to its derivative, in this case, three plus\n

finding a derivative. Anti differentiating,\n

other direction, from a derivative to a function\n

For example, if g prime of x is 3x squared,\n

original function be? Well, g of x could be\n

3x squared. Or it could also be g of x equals\n

equals x cubed plus any constant, where I\n

That's because the derivative of a constant\n

constant is just going to be 3x squared, no\n

F of X is called an antiderivative of lowercase\n

capital F prime of X is equal to lowercase\n

we can think of little f as being the derivative\n

x cubed is an antiderivative of 3x squared.\n

C is also an antiderivative of 3x squared.\n

sometimes referred to as a general antiderivative,\n

for the function 3x squared. But could there\n

whose derivative is 3x squared. In fact, there\n

this intuitively, is if you have two functions\n

two runners in a race that always speed up\n

one of those runners starts ahead of the other,\n

stay exactly the same. That's the vertical\n

the constant C, that separates one antiderivative\n

plus C. And in general, if capital F of X\n

then all other anti derivatives can be written\n

constancy. A more rigorous justification of\n

theorem, as I'll do in a separate video. If\n

functions, then it's pretty easy to get some\n

of one is x. Since the derivative of x is\n

we can add a constant C, the antiderivative\n

take the derivative of x squared over two,\n

with the two in the denominator, leaving me\n

by adding a constant C. More generally, the\n

not equal to negative one is given by x to\n

a constant C. I can check this by taking the\n

one. The n here is just a constant. So using\n

the n divided by n plus one that yields x\n

think of this rule as the power rule for anti\n

to the power rule for differentiating. Now,\n

one. Notice that we'd be dividing by zero\n

case when n equals negative one separately,\n

x, we recognize that the antiderivative of\n

of x plus C. Since the derivative of ln of\n

pause the video and see how many more anti\n

You should get all of these formulas based\n

notice that the antiderivative of sine x is\n

derivative of cosine x is negative sine x.\n

going to call the constant a instead of C,\n

If I want the antiderivative of A times x\n

the antiderivative of x to the n, which is\n

constant say that's because when I take the\n

I can just pull the constant out. More generally,\n

function, little f of x is just going to equal\n

x, which I'll denote with capital F of X,\n

of x plus g of x is capital F of X, plus capital\n

G are the antiderivative of lowercase F and\n

of a sum is equal to the sum of the derivatives.\n

antiderivative for f of x equals five over\n

the square root of x. First, I'm going to\n

plus x squared minus one half times x to the\n

of one over one plus x squared is arctangent\n

the antiderivative of x to the minus one half,\n

one half plus one is positive one half, and\n

multiplication rules, I can just multiply\n

capital F of X, I've have to remember the\n

simplify a little bit by canceling these one\n

arc tan of x minus a squared of x plus C.\n

and build a table of anti derivatives based\n

we'll solve problems where we're given an\n

And we're given an initial condition, something\n

find the function f of x. In this first example,\n

times sine x. and g of two pi is five, we\n

of e to the x minus three sine x. So g of\n

of x plus a constant C. That's because the\n

derivative of cosine x is minus sine x, and\n

Now I need to find the value of the constant\n

If I plug in two pi for x, I get e to the\n

C, and that needs to equal five. Since cosine\n

pi plus three plus c equals five. And so C\n

my function g of x is equal to either the\n

the two pi. In this example, we're given the\n

initial conditions, f of one is zero, and\nf of zero is two.

To start, I'm going to rewrite f double prime\n

I get the square root of x times x minus the\n

exponents, we get x to the three halves minus\n

find f prime of x, which is the antiderivative\n

the power rule for anti derivatives, I raised\n

five halves, and then divide by the new exponent\n

one half by one to get one half and divide\n

on a constant see, let me simplify a little\n

multiply by two fifths, and instead of dividing\n

an expression for f prime of x, but I need\n

of f prime, and so all anti differentiate\n

the seven halves over seven halves minus two\n

the antiderivative of a constant C is C times\n

After simplifying a bit, I'm ready to use\n

my constant C and D. When I plug in zero for\n

D, so d has to equal two. So I can rewrite\n

my second condition says that f of one equals\n

fifths minus four thirds plus c plus two,\n

c is negative two, minus 430 fifths, plus\n

to 100, and fifths. If we plug that in for\n

finishes the problem. In this final example,\n

to make them up ourself, we're told that we're\n

30 meters, with throw a tomato up in the air\n

The tomato then falls down to the ground due\n

takes and what its velocity is at impact.\n

is negative 9.8 meters per second squared.\n

figure if we're working in units of feet,\n

negative sign is because gravity is pulling\n

direction. We're also given the initial condition,\n

second. That's a positive velocity because\n

that the initial position s of zero is 30\n

acceleration is negative 9.8. In other words,\n

S prime of t is negative 9.8 t plus a constant\n

velocity, I know that s prime of zero is 20.\n

9.8 times zero plus c one that has to equal\n

Substituting in 20 for C one, I can rewrite\n

of S prime. And that's going to be negative\n

Using my second initial condition, f of zero\n

an expression that equals 30. Since all the\n

me that C two is 30. And so I can find my\n

Now I want to find out how long it takes the\n

to be the time when s of t equals zero. Setting\n

can use the quadratic formula to solve for\n

1.17, or 5.25. The negative time doesn't make\n

time of impact of 5.25 seconds. Now to find\n

that time into my velocity equation. In other\n

of 5.25, is 9.8 times 5.25, plus 20, which\n

probably enough to squash the tomato. And\n

derivatives using initial conditions. In this\n

show that the antiderivative zero has to be\n

of the same function have to differ by a constant.\n

if f of x is one antiderivative of a function,\n

of that same function can be written in the\n

C. In other words, any two antiderivative\n

To prove this fact, let's first note that\n

x is equal to zero on an interval than the\n

some constant C. This statement follows from\n

theorem tells us that for any x one and x\n

between x one and x two is equal to the derivative\n

x two. But by assumption, g prime is zero\n

x three must be equal to zero. This means\n

of x one has to equal zero. In other words,\n

true for any x one and x two. So all the values\n

The second observation that I want to make\n

which have the same derivative, then g one\n

constant C. This statement follows from the\n

of x is equal to g two prime of x, then g\n

equal zero. But that means if I look at the\n

take its derivative, that has to equal zero,\n

difference of the derivatives. Now our previous\n

of a function is zero, the function must be\n

g two of x equals C for some constant C, which\n

x plus C, which is what we wanted to prove.\n

the same derivative have to differ by a constant\n

antiderivative of a function than any other\n

F of x plus C. This concludes the proof that\n

must differ by a constant. This video will\n

notation used to write a song.

In this expression, using the greek capital\n

The number one is called the lower limit of\n

number five is called the upper limit of summation,\n

by summing up to the I, for all values of\n

stepping through the integers. In other words,\n

the one. And then we add to it two to the\n

two to the five. If we do the arithmetic,\n

our index is J, and we start with J equals\n

stepping through integer values. So we have\n

six, and then our last term is one seven.\n

given a psalm like this one, it can be handy\n

more compact. But to do so, we have to look\n

case, the terms all differ by three. So I\n

and 12, as being six plus twice, three, and\n

so on. In fact, we can even think of six itself\n

this pattern. Now we can write the sum as\n

from zero to four. Here, we're thinking of\n

Now, there are other ways to write this sum\n

notice that each of the terms is a multiple\n

two, nine is three times three, and so on.\n

three times n, where n ranges from two to\n

index doesn't matter at all. For example,\n

two to six of three times k. Here, k and n\n

for a moment and try to write this next example\n

are powers of two, we could write the denominators\n

five. The numerators are one less than the\n

as two to the i minus one was we're adding\n

from two to five. In this video, we reviewed\n

writing sums. In this video, we'll approximate\n

this will introduce the idea of an integral.\n

this curve y equals x squared in between x\n

There's several ways to do this. For example,\n

of each rectangle is as tall as the curve.\n

we could line up our rectangles, so the left\n

We'll call this use Left endpoints. Notice\n

is degenerate and has height zero. If we're\n

then the base of each rectangle has size one\n

given by the value of our function, y equals\n

So for example, the area of the first rectangle\n

five times 0.5 squared. The 0.5 squared comes\n

0.5 to get the height. Similarly, the area\n

base times height basis, still point five,\n

or one. If we continue like this, and add\n

rectangles is given by this expression. Notice\n

rectangle, each rectangle has base of point\n

we can write this in sigma notation as the\n

I ranges from one to six. This works because\n

of point five. The first one is point five\n

five times six. Now if we compute the sum,\n

from the picture, that the sum of areas of\n

under the curve, we can do the same sort of\n

endpoints, and we'll get an under estimate\n

to try it for yourself before going on with\n

rectangle has area zero, the second rectangle\n

height of 0.5 squared. And if we compute all\n

expression to the previous one, only this\n

the height of our last rectangle. One way\n

starting with i equals one for the first rectangle\n

And then for the height, we use 0.5 times\nI minus one

This works because when I is one, i minus\n

zero like we should. And when I is six, we\n

is point five times five, or 2.5, just like\n

get an answer of 55 eighths, which is equal\n

true area under the curve. Now there's a big\n

to get tighter bounds on the area. One way\n

example, 12 rectangles instead of six. Again,\n

gives us an over estimate of area in this\n

gives us an under estimate. The area of the\n

the height, and the base is going to be in\n

by the functions value on the right endpoint,\n

going to be 0.25 times I and the function\n

be given by point two, five i squared. The\n

by the sum from i equals one to 12 for the\n

If we work out that sum, it comes to 10.156.\n

endpoints. Now the area of the eye, the rectangle,\n

two, five. And now the height is given by\n

So that's going to be 0.25 times I minus one.\n

y equals x squared is giving the height of\n

together is going to be the sum from the first\n

That works out to 7.906. So we're honing in\n

somewhere between about eight and about 10,\n

of area by using more and more rectangles.\n

then our area of all rectangles using right\n

from i equals one to 100 of the basis times\n

will be 100 of the length here from zero to\n

over 100. The eyes right endpoint, which I'll\n

100 times I, since you get to the right eye\n

of width three one hundredths. Therefore,\n

is right endpoint squared, or three one hundredths\n

areas as sigma from i equals one to 100 of\n

times i squared. The formula using left endpoints\n

to be three one hundredths, times I minus\n

we only have to travel through i minus one\n

the rectangle. So our area for the left endpoints\n

three over 100 times three over 100 times\n

out to be 9.1435 and 8.8654. At this point,\n

our exact area under the curve is probably\n

area for sure, let's do this process of dividing\n

we'll just use an rectangles where n is some\n

of length three into n little pieces, the\n

the base of each rectangle, we'll be given\n

a little tiny bit of x. Now the right endpoint,\n

we have to travel through I rectangles, each\n

right endpoint. So our height, H sub i is\n

endpoint. We can work out similar expressions\n

Our estimate of area using right endpoints\n

three over n times three I over n squared.\n

sum from i equals one to n of three over n\n

more rectangles we use, in other words, the\n

area will be to our exact area under the curve.\n

the limit the limit as n goes to infinity\n

there are really two possible limits, we could\n

as the picture suggests, these two limits\n

there are other options between sides using\n

for example, use the midpoints of our intervals\n

limit should also end up as the same thing.\n

values of x equals one and x equals three.\n

called a Riemann sum. I'll stick with the\n

the exact area, we have to evaluate this limit,\n

Since three and n don't involve the index\n

sign. I'll clean this up a little bit. Now\n

of the first n squares of the integers is\n

we can check that formula for a few values\n

summing out one squared plus two squared,\n

two plus one times four plus one over six\n

If we use this formula, in our limit calculation,\n

nine halves, by dividing 27 by six, we can\n

two n plus one over n squared. I'm going to\n

I have the limit of a rational expression,\n

is going to be two n squared, the highest\n

so that's going to be a limit of two, multiply\n

of nine, just like I expected from the previous\n

did successfully find the area under the curve,\n

the area under a curve by taking the limit\n

of the area of the rectangles, which is given\n

times the heights of the rectangles. The basis\n

are given by the function value on the left\n

point in the interval. For our purposes, f\n

called a Riemann sum, can be used more generally,\n

function. In previous sections, we thought\n

and we've computed it as a number. In this\n

as a function of the bounds of integration.\n

theorem relating the derivative and the integral.\n

let g of x be the integral from one to x of\n

my integrand Here, just to distinguish it\n

of integration, this expression just means\n

on the x axis. I'll call geovax, the accumulated\n

x, measures how much net area has accumulated.\n

of x. g of one is the integral from one to\n

the bounds of integration here the same, g\n

f of t, dt. That's the net area from one to\n

is the integral from one to three. Now, we've\n

an additional one unit up here from this triangle,\n

with some additional area tacked on the additional

area measures three units. So g of four is\n

When we go from g of four to g of five, we\n

is nine. As we go from g of five to g of six,\n

f is now below the x axis. So here I've accumulated\n

g of six is one less than g of five. In other\n

Since we accumulate three more units of negative\n

one to zero of f of t dt, I'm going to rewrite\n

zero to one of f of t dt. Since there are\n

of zero is negative to apply all these values\n

the dots to get an idea of what g of x looks\n

g prime of x. We know that g prime of x is\n

g of x is increasing, wherever we're adding\n

So we have that g prime of x is positive,\n

g prime of x is negative, where g of x is\n

on negative area because f of x is negative.\n

where f of x is negative. Also, g prime is\n

At that instant, we're not adding on any positive\nor negative

If we look a little closer, we can see the\n

on the height of f of x. When f of x is tall,\n

While when f of x is low or small, we're adding\n

of G. In other words, g prime of x is behaving\n

And in fact, it turns out that g prime of\n

of the fundamental theorem of calculus.

The Fundamental Theorem of Calculus Part One\n

on the closed interval from a to b, then for\n

x, the integral from a to x of f of t dt is\n

on the inside of this interval, and for Furthermore,\n

in the previous example. The proof of this\n

and can be found in a later video. For now,\n

First, let's find the derivative with respect\n

square root of t squared plus three dt. The\n

this expression here thought of as a function\n

is just the integrand function evaluated on\n

work here at all. To evaluate the derivative,\n

the derivative and the second expression is\n

it might seem odd that these two expressions\n

both cases, we're taking the derivative of\n

at which area accumulates doesn't depend on\n

it doesn't depend on where we start counting,\n

For this third example, remember that the\n

as the negative of the integral from four\n

from four to x. and applying the fundamental\n

the square root of x squared plus three, it\n

answer for this example. When we're integrating\n

actually decreases. So our accumulated area\n

This last example is more complicated, because\n

we have a function of x sine of x, we can\n

and the accumulated area function as being\n

In general, the chain rule says that we have\n

of u of x, then that's the same thing as the\n

at U times the derivative of the inside function\n

rule to our accumulated area function, where\n

respect to x of the integral from four to\n

DT can be written as the derivative by two\n

to you have the integrand times the derivative\n

just sine x, we can apply the fundamental\n

derivative. By just plugging in you for T,\n

derivative of sine x, of course, is just cosine\n

entirely in terms of x, we're going to rewrite\n

plus three times cosine x, or just the square\n

of x. We could have gotten this answer more\n

sine of x in where we saw the T here in the\n

by the derivative of sine x due to the chain\nrule.

This video introduced the fundamental theorem\n

of the integral of a function is just the\n

derivative and does the process of taking\n

closely related. inverse operations. This\n

theorem of calculus. Another way of relating\n

fundamental theorem of calculus says that\n

interval a b, then the integral from a to\n

minus capital F of A, where capital F is any\n

F is a function whose derivative is lowercase\n

theorem of calculus follows directly from\n

video. But here, I just want to make a few\n

we think of f of x as the derivative of capital\n

of the derivative is equal to the original\n

want to comment on the phrasing any antiderivative.\n

for lowercase F. We know that any two anti\n

that g of x has to equal capital F of X plus\n

g of a, that's going to be the same thing\n

since this constant sees subtract out to cancel\n

this difference is the same value, no matter\n

And that's why we can say that capital F can\n

theorem of calculus is super useful, because\n

finding anti derivatives and evaluating them.\n

easy. Computing integrals, using the Riemann\n

of the fundamental theorem of calculus, we\n

and tedious computations, we've involving\n

to do to evaluate an integral is find an antiderivative,\n

in some examples. In this first example, the\n

the antiderivative of negative four over x\n

We could add a plus C to make it a general\n

The fundamental theorem says that we can use\n

the simplest one, where c equals zero. Now\n

the endpoints of negative one and negative\n

line with a negative one at the bottom and\n

In other words, the notation capital F

between A and B means capital F of b minus\n

for our antiderivative here. So now we just\n

what we get when we plug in negative one for\n

to write the antiderivative of one over x\n

because ln of the absolute value of five,\n

whereas ln of negative five would not exist.\n

I get negative 125 minus four ln five minus\n

of one is zero, this becomes negative 124\n

130 point 438. In this next example, we need\n

y squared minus y plus one over the square\n

separately of the numerator and the denominator,\n

works for differentiation. So it doesn't work\n

Instead, let's try to simplify this expression\n

take the antiderivative of. So I'm going to\n

power. And dividing by y to the one half is\n

negative one half distributed, distributing\n

halves minus y to the one half plus y to the\n

can take the antiderivative of just using\n

halves, becomes y to the five halves by adding\n

Now here, I get y to the three halves divided\n

half plus one is positive one half, I need\n

me simplify a little bit. And now I'll substitute\n

the same thing as four to the one half raised\n

power, or 32. Similarly, for the three halves\n

cubed, or eight, and four to the one half\n

one. And after some arithmetic, I get an answer\nof 146 15th.

The Fundamental Theorem of Calculus, part\n

capital F with continuous derivative, the\n

original function evaluated on the bounds\n

both parts of the fundamental theorem of calculus.\n

of calculus says that if f of x is a continuous\n

as the integral from a constant A to the variable\n

derivative equal to the original function

To prove this theorem, let's start with the\n

g prime of x, by definition, is the limit\n

of x over h. Now g of x is defined as an integral\n

be the integral from a to x plus h, just plugging\n

By properties of integrals. The integral from\n

x is just the integral from x to x plus h.\n

h can be closely approximated by a skinny\n

And so this limit is approximately the limit\n

which is just f of x. But let's make this\n

capital M be the maximum value that f of x\n

lowercase m be the minimum value achieved.\n

of the interval from x to x plus h, but they\n

But we know that f of x does have to have\n

it's a continuous function by assumption on\n

of f of t dt from x to x plus h has to be\n

bigger than a to lowercase m times h. This\n

can be verified visually by comparing this\n

which has area lowercase m times h, and comparing\n

has area capital M times h equivalently, the\n

by h has to be less than or equal to capital\n

the intermediate value theorem, which holds\n

intermediate value that lies between the minimum\n

as f of c for some C in the interval. Therefore,\n

expression above by just simply the value\n

The value of C here depends on x and h. But\n

closer to x. And since f is continuous, this\n

We've now proved the first part of the fundamental\n

g exists and equals f of x. The second part\n

that if f is continuous, then the integral\n

of lowercase F, which I'll denote by capital\n

evaluated today. Part Two of the fundamental\n

let g of x be defined as the integral from\n

fundamental theorem of calculus tells us that\n

of x. In other words, capital G is an antiderivative\n

A is by definition, the integral from a to\n

a to a of f of t dt. The second integral is\n

identical. So part two of the fundamental\n

But the theorem is supposed to be true for\n

be any antiderivative of lowercase F, we know\n

of X plus some constant since any two antiderivative\n

and therefore, capital F of b minus capital\n

C minus capital G of A plus C. The constant\n

b minus capital G of A, which we already saw\n

f of t dt. So the left side of this equation\n

theorem of calculus Part two is proved for\n

of the fundamental theorem of calculus. This\n

evaluating integrals, also known as u substitution.\n

to x sine of x squared dx. Now sine of x squared\n

the function x squared. And notice that the\n

is sitting right here and the integrand. I'm\n

squared, and then I'll write d u is equal\n

find d u, I take the derivative have X squared\n

I can then rewrite the integrand as sine of\n

this substitution, I can integrate, because\n

cosine of u. And I'll add on the constant\n

problem was in terms of x, and now I've got\n

back in since u is equal to x squared, I can\n

To verify that this final answer is correct,\n

we started with, let's take the derivative\n

If we take the derivative of negative cosine\n

of a constant is zero, so we have the derivative\n

the inside function x squared times the derivative\n

And we do in fact, get back to the integrand\n

chain rule when taking the derivative to check\n

use substitution. When looking for what to\n

chunk that's in the integrand, whose derivative\n

to just have a constant multiple of the derivative\n

we might use the chunk one plus 3x squared,\n

is six times x. And even though six times\n

have a factor of x in the numerator, that's\n

of 6x. So let's write out d u, that's going\n

rewrite this as x dx is equal to one six d\n

substitute one six d u for x dx. And then\n

becomes u. I can rewrite this as one six times\n

that the antiderivative of one over u is ln\n

for you, I get a final answer of one six ln\n

say, the absolute value signs are not really\n

3x squared is always positive. As our next\n

the 7x dx. one chunk with us here is u equals\n

And so we have dx is equal to 1/7 do substituting\n

1/7 d u, I can pull the 1/7 out and integrate\n

substituting in for back for 7x, I get e to\n

video to check that these two answers are\n

that you use the chain rule each time. Next,\n

the integral from E to E squared of ln x over\n

is the derivative of ln x, that's one of our\n

you than say setting u equal to x from the\n

dx. And when we did the substitution, nothing\n

we need to deal with the bounds of integration\n

worry about them now or worry about them later.\n

first. Our bounds of integration E and E squared\n

in our integral from x to you, we need to\n

of x to values of u also. Now, when x is equal\n

using this equation. Similarly, when x is\n

which is two. So as I rewrite my integral,\n

two. And now my ln x becomes my u and my dx\n

to grow of UD u is equal to use squared over\n

of u equals two and u equals one to get two\n

which is one half. Notice that when we did\n

to get back to our variable x, we stayed in\n

of dealing with the bounds of integration\n

to the beginning of the problem. We're just\n

equals one over x dx. Instead of substituting\n

to temporarily ignore them and just evaluate\n

I can substitute in as you times do, we can\n

Normally we'd have a plus c constant. But\n

definite integral, we don't really need the\n

indefinite integrals, I'm going to get back\n

for you U is ln of x. So I square that and\n

my original bounds of integration, those bounds\n

in those bounds, I get ln of E squared quantity\n

two, which evaluates to two squared over two\n

This video gave some examples of use substitution\n

in examples like this one, where there's a\n

or at least a constant multiple of its derivative\n

You've already seen how u substitution works\n

why it works. u substitution is based on the\n

we take the derivative of a function, capital\n

get the derivative of capital F evaluated\n

If we just write that equation in the opposite\n

f prime of g of x times g prime of x can all\n

function, f of g of x. Now if I take the integral\n

to x, on the right side, I'm taking the integral

Well, the integral or antiderivative of a\n

capital f of g of x plus C. Now when we do\n

this equation down. We are seeing an expression\n

of x dx, we're recognizing you as g of x and\n

this expression as the integral of capital\n

just capital F of u plus C. And then we're\n

f of g of x plus C, the beginning and end\n

left side and right side of our chain rule\n

to integrate, you can thank this chain rule\n

the idea of an average value of a function.\n

we just add the numbers up and divide by n,\n

we write the sum from i equals one to n of\n

value of a continuous function is a little\n

infinitely many values on an interval from\n

of the function by sampling it at a finite\n

them x one through x n. And let's assume that\n

then the average value of f at these sample\n

divided by n, the number of values are in\n

one to n of f of x i all divided by n. This\n

we're just using n sample points. But the\n

sample points n gets bigger and bigger. So\n

n goes to infinity of the sample average.\n

sum. So I need to get delta x in there. So\n

bottom by delta x. And notice that n times\n

b minus a. Now as the number of sample points\n

them goes to zero. So I can rewrite my limit\n

sum of FX II times delta x divided by b minus\n

numerator is just the integral from a to b\n

the function is given by the integral on the\n

of the interval. Notice the similarity between\n

and the formula for the average value of a\n

corresponds to the summation sign for the\n

B minus A for the function corresponds to\n

Now let's work an example. For the function\n

interval from two to five, we know that the\n

from two to five of one over one minus 5x\n

I'm going to use use of the tuition to integrate,\n

So d u is negative five dx. In other words,\n

bounds of integration, when x is equal to\ntwo, u is equal to

one minus five times two, which is negative\n

to negative 24. substituting into my integral,\n

24 of one over u times negative 1/5. Do and\n

is same as multiplying by 1/3. And as I integrate,\n

then take the integral of one over u, that's\n

between negative 24 and negative nine. The\n

they prevent me from trying to take the natural\n

get negative 1/15 times ln of 24 minus ln\n

and get negative 1/15 ln of 24 over nine,\n

as a decimal, that's approximately negative\n

Now my next question is, does g ever achieve\n

a number c in the interval from two to five\n

one way to find out is just to set GFC equal\n

one over one minus five c equal to negative\n

for C. There are lots of ways to solve this\n

of both sides, subtract one from both sides\n

to three over ln of eight thirds, plus 1/5,\n

does lie inside the interval from two to five.\n

its average value over the interval. But in\n

Gs average value has to lie somewhere between\n

interval. And since G is continuous on the\n

every value that lies in between as minimum\n

The same argument shows that for any continuous\n

value on an interval. And this is known as\n

for any continuous function f of x on an interval\n

number c, between A and B, such that f of\n

f of c equals the integral from A b of f of\n

the definition of an average value of a function,\n

If we rewrite the formula for average value\n

for average value, the area of the box with\n

area under the curve. This video gives two\n

the mean value theorem for integrals says\n

on an interval from a to b, there's some number\n

to the average value of f. The first proof\n

value theorem. Recall that the intermediate\n

function f, defined on an interval, which\n

l in between f of x one and f of x two, then\n

x one and x two. Keeping in mind the intermediate\n

to the mean value theorem for integrals. Now\n

be constant on the interval from a to b. But\n

for integrals holds easily, because f AV is\n

to f OC for any c between A and B. So let's\n

continuous function on a closed interval has\n

which I'll call little m, and big M. Now we\n

has to be between its maximum value and its\n

consider the fact that all of us values on\n

little m. And if we integrate this inequality

We get little m times b minus a is less than\n

or equal to big M times b minus a. Notice\n

just integrating a constant. Now if I divide\n

little m is less than or equal to the average\n

as I wanted. Now, I just need to apply the\n

as my number L and little m and big M as my\n

value theorem says that F average is achieved\n

x two. And therefore, for some C in my interval\n

for integrals. Now I'm going to give a second\n

And this time, it's going to be as a corollary\n

functions. Recall that the mean value theorem\n

and differentiable on the interior of that\n

interval, such that the derivative of g at\n

G, across the whole interval from a to b.\n

in mind, and turn our attention back to the\n

to define a function g of x to be the integral\n

given to us in the statement of the mean value\n

is just the integral from a to a, which is\n

to b of our function. Now, by the fundamental\n

continuous and differentiable on the interval\n

And by the mean value theorem for functions,\n

b minus g of a over b minus a, for some numbers,\n

the three facts above, into our equation below,\n

a to b of f of t dt minus zero over b minus\n

wanted to reach. This shows that the mean\n

mean value theorem for functions where our\n

the second proof of the main value theorem\n

value theorem for integrals in two different\n

of calculus along the way.