# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 3

Find d/dπ₯ β«_(5)^(π₯β΄) ln(7 + π‘Β³) dπ‘.

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### Video Transcript

Find d by dπ₯ of the integral evaluated between five and π₯ to the fourth power of the natural log of seven plus π‘ cubed with respect to π‘.

Now, as tempting as it may be, we absolutely do not want to evaluate the integral here. Instead, weβ²re going to recall the second fundamental theorem of calculus. This says that if π is a continuous function and π is a constant, then the integral evaluated between π and π₯ of π of π‘ with respect to π‘ is equal to capital πΉ of π₯ minus capital πΉ of π, where capital πΉ prime of π₯ is equal to the function of π₯. And remember, capital πΉ is the antiderivative of the function. So how does this help us?

Weβ²re going to differentiate both sides of the equation with respect to π₯. And we see that we can separate our derivative into d by dπ₯ of capital πΉ of π₯ minus d by dπ₯ of capital πΉ of π. Remember, capital πΉ of π will give us a constant. This means itβ²s independent of π₯. And so, its derivative is going to be zero. So weβre left with the derivative of capital πΉ of π₯ with respect to π₯ or capital πΉ prime of π₯. Earlier though, we said that capital πΉ prime of π₯ was equal to π of π₯. And we therefore see that the derivative with respect to π₯ of the integral evaluated between π and π₯ of π of π‘ with respect to π‘ is equal to π of π₯.

Letβ²s see how we can use this to evaluate the derivative with respect to π₯ of the integral between five and π₯ to the fourth power of the natural log of seven plus π‘ cubed with respect to π‘. We can say that in our example, π of π‘ is equal to the natural log of seven plus π‘ cubed. And we use the rules we developed earlier. And we see that the derivative with respect to π₯ is equal to d by dπ₯ of capital πΉ of π₯ to the fourth power minus capital πΉ of five. We split our derivative up. And then we recall that the derivative of capital πΉ of five with respect to π₯ is going to be equal to zero, since itβ²s going to be independent of π₯.

And so, we see that the derivative with respect to π₯ of the integral evaluated between five and π₯ to the fourth power of the natural log of seven plus π‘ cubed with respect to π‘ is the derivative of capital πΉ of π₯ to the fourth power with respect to π₯. Weβ²re going to need to be really careful here when evaluating d by dπ₯ of capital πΉ of π₯ to the power of four. Weβ²re going to be differentiating a function of a function, or a composite function. So weβ²re going to need to use the chain rule.

This says that if π¦ is a function in π’ and π’ itself is a function in π₯, then the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. So this means that the derivative of capital πΉ of π₯ to the power of four with respect to π₯ is equal to the derivative of capital πΉ of π₯ to the power of four, which is just πΉ prime of π₯ to the power of four multiplied by the derivative of π₯ to the power of four with respect to π₯. Thatβs four π₯ cubed.

Now, we said earlier that πΉ prime of π₯ was equal to π of π₯. So we can say that πΉ prime of π₯ to the power of four is equal to π of π₯ to the power of four. Since we said that π of π‘ was equal to the natural log of seven plus π‘ cubed, we know that π of π₯ to the fourth power is equal to the natural log of seven plus π₯ to the fourth power cubed times four π₯ cubed. π₯ to the fourth power cubed is π₯ to the power of 12. And we can see that the derivative with respect to π₯ of the integral evaluated between five and π₯ to the fourth power of the natural log of seven plus π‘ cubed with respect to π‘ is equal to four π₯ cubed times the natural log of seven plus π₯ to the power of 12.