# Question Video: Finding the Derivative of a Function Defined by an Integral Mathematics • Higher Education

Use the fundamental theorem of calculus to find the derivative of the function π¦ = β«_(4)^(5π₯ +3) (2π‘/(2 + π‘β΅)) dπ‘.

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### Video Transcript

Use the fundamental theorem of calculus to find the derivative of the function π¦ is equal to the integral from four to five π₯ plus three of two π‘ divided by two plus π‘ to the fifth power with respect to π‘.

We need to find the derivative of a function π¦. And we might not be sure what to differentiate this with respect to because we see that there are two variables, π‘ and π₯. However, we need to notice that π‘ is a dummy variable. Remember, the last step when we use a definite integral is to substitute our limits of integration into our antiderivative. So π¦ is a function of π₯, and we need to differentiate π¦ with respect to π₯. And weβre told to do this by using the fundamental theorem of calculus.

So letβs start by recalling the fundamental theorem of calculus. This says if lowercase π is a continuous function on a closed interval from π to π and capital πΉ of π₯ is the definite integral from π to π₯ of lowercase π of π‘ with respect to π‘, then we can find an expression for capital π prime of π₯. Thatβs the derivative of the integral from π to π₯ of lowercase π of π‘ with respect to π‘ with respect to π₯. And itβs equal to lowercase π of π₯ for all values of π₯ in the open interval from π to π.

So the fundamental theorem of calculus gives us a way of differentiating complicated-looking integrals where π₯ is one of the limits of integration. And we can see this is the situation we have here. For example, our function capital πΉ of π₯ will be π¦. We see the lower limit of integration is four. That will be our value of π. Our integrand two π‘ divided by two plus π‘ to the fifth power will be our function lowercase π of π‘. But then we see a problem. Our upper limit of integration is not π₯. We have a function in π₯, five π₯ plus three.

But we can get around this problem. Letβs start by setting π’ equal to five π₯ plus three. So, in our definition for π¦, weβll substitute π’ is equal to five π₯ plus three into our upper limit. This means that π¦ is equal to the definite integral from four to π’ of two π‘ divided by two plus π‘ to the fifth power. But remember, weβre still trying to differentiate π¦ with respect to π₯. We want to find an expression for the derivative of π¦ with respect to π₯, so weβll differentiate both sides of this equation with respect to π₯.

This gives us π¦ prime of π₯ is equal to d by dπ₯ of the integral from four to π’ of two π‘ divided by two plus π‘ to the fifth power with respect to π‘. And using exactly the same argument we used at the start of this video, we can see that this integral is a function in π’. So weβre trying to differentiate a function of π’ with respect to π₯. But remember, π’ is a function in π₯, so we can do this by using the chain rule. So letβs recall what the chain rule tells us.

The chain rule tells us if π¦ is a function in π’ and π’ is a function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. And this is exactly the situation we have here. π¦ is a function in π’ and π’ is a function in π₯. So we can find an expression for dπ¦ by dπ₯ by finding dπ¦ by dπ’ and multiplying this by dπ’ by dπ₯. So by using the chain rule, we have that π¦ prime of π₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.

And now, we can just apply the fundamental theorem of calculus to evaluate this first derivative. Weβre differentiating our integral with respect to π’. The lower limit of integration is a constant, and the upper limit of integration is π’, and our integrand is a rational function. This means itβs continuous across its entire domain. And the only time a rational function is not defined is when its denominator is equal to zero. And of course we can solve this. This will be when π‘ is equal to negative the fifth root of two. Therefore, we can use the fundamental theorem of calculus to evaluate this derivative. It will be our integrand evaluated at π’.

So by using the fundamental theorem of calculus, we have dπ¦ by dπ’ is equal to two π’ divided by two plus π’ to the fifth power. But remember, we still need to multiply this by dπ’ by dπ₯. And remember, π’ is a function in π₯. Itβs five π₯ plus three. This is a linear function, so its derivative with respect to π₯ will be the coefficient of π₯, which is five. So we have now shown that π¦ prime of π₯ will be equal to two π’ divided by two plus π’ to the fifth power multiplied by five.

Of course, we can simplify this. Five multiplied by two simplifies to give us 10. But weβre not done yet. Remember, π¦ prime of π₯ is a function in π₯, so we should use our substitution π’ is equal to five π₯ plus three to rewrite our answer in terms of π₯. And so, using our substitution π’ is equal to five π₯ plus three, we get our final answer of π¦ prime is equal to 10 times five π₯ plus three all divided by two plus five π₯ plus three all raised to the fifth power.