Original subtitles

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Today we're going to talk about how to find f of x even the functions third derivative.

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Complete this problem.

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We'll work backwards taking the integral of the third derivative to get the second derivative.

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The integral of the second derivative they get the first derivative and then the integral of the first

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derivative to find the original function.

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In this particular problem we've been asked to find f of x if f triple prime or the third derivative

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of x is equal to x minus the square root of x.

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So in order to find our way from F triple prime or the third derivative X back to f of x we're going

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to have to work our way backwards.

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So we'll start with the third derivative of X and we'll we'll try to find the second derivative X on

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we found it will find the first derivative and then use the first derivative to find the original function.

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So we'll start with the third derivative of X and in order to find the second derivative of x f double

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prime we'll need to take the integral or the antiderivative of the third derivative.

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So X minus the square root of x and whenever you have integral notation like this you're going to need

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a DX corresponding notation to go with it.

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So let's go ahead and make a change in this integral to make it easier on ourselves.

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We're going to be taking the integral of x.

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I want to convert the square root of x here to x to the one half which is the same thing because now

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we can use power rule to easily take integral of this term.

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So to take the integral or remember that we're going to use simple power rule here.

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We have X to the first power x to the one that take the integral we'll add 1 to the exponent.

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We'll get one plus one equals two and we'll divide the coefficient which right now is just 1 divided

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by the new exponent which we already found to be 2.

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So that's how we use power rule to take the integral of simple polynomial terms like this.

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So now we need to take the integral of negative x to the one half in order to do that.

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Add 1 to the exponent.

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So one half plus one will give us X to the three halves and then we'll divide the coefficient which

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is currently 1 by the new exponent so divided by three halves.

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When we simplify this we'll get 1 1/2 x squared minus two thirds.

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X to the three halves but then keep in mind that when ever we take an integral we always have to add

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C. or the constant integration.

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So we need to add c to account for the constant of integration here.

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And of course that would carry down into our simplification.

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So what we solve for here is the second derivative of f of x.

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Now we need to go ahead and find the first derivative.

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So the first derivative will be f prime of X and that will be equal to the integral of the second derivative.

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So 1 1/2 x squared minus two thirds X to the three halves class C D.

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So again we'll use power rule to take the integral of this term by term.

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We have x squared year.

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Add one to the exponent two plus one gives us three and then we divide the coefficient one half by three.

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One half divided by three is one sixth.

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So we get one 6 there.

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Now we have negative two thirds X to the three halves that take the integral will add 1 to the exponent

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three halves plus one will give us five halves.

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So X to the five halves and then we'll divide negative two thirds by five halves.

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The new exponent but negative two thirds divided by five halves is the same thing as negative two thirds

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times two fifths.

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Right.

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Remember instead of dividing by a fraction we can multiply by its inverse So this is negative two thirds

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times two fifths.

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And then when we take the integral of C we'll just get C X because remember that C is just a constant.

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And this is basically c times x to the zero power x to the zero and anything raised to the zero is 1.

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So essentially we haven't changed this at all because we're just saying See times 1 which is see this

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now follows though the same rules as the other two terms so since we have X to the 0 will add 1 to the

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exponent and get X to the first power and then divide the coefficients C at a new exponent 1.

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But of course because we've integrated We also want to add another concept since we've already used

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SI to account for a constant we add D instead.

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Now we just need to simplify this as much as we can so we'll get one sixth X to the third we have negative

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two thirds times two fifths.

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So we'll get minus four over 15 x to the five halves plus C X plus D.

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And that is our first derivative.

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Now in order to finally get back to our original function f of x we will take the integral of our answer

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there so our backs will be the integral of one sixth next to the third minus for over 15 x to the five

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halves plus X plus D.

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And as always we'll have our the X here to take the integral we'll do the same thing that we've been

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doing well add 1 to the exponent to x to the third becomes X to the fourth.

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And now we have 1 6 divided by 24 which will be 1 24th taking the integral of this X to the five halves

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will add 1 to the exponent 5 habes plus 1 gives us seven halves.

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So X to the seven hands will get negative for over 15.

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Instead of dividing by seven halves will multiply by the inverse to sevens and we get Plus we add 1

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to the exponent.

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This is essentially X to the first so get x squared.

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And we'll divide the coefficients C by the new exponent which is two and then same thing here.

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The this is essentially the times x to the 0.

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We add one to the explant and we get access to the first power.

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We divide the by the new exponent.

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So we get the over 1.

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But because we're integrating Of course we have to add another constant of integration this time we'll

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call it e.

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So when we simplify this we'll get our final answer for f of x which is just going to be 1 over 24 x

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to the fourth negative for 15 times to sevenths will be negative 8 over 1 0 5 x 2 the 7 hands plus C

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over to x squared plus the X plus E.