# 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 𝑦 = ∫_(4)^(5𝑥 +3) (2𝑡/(2 + 𝑡⁵)) d𝑡.

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### Video Transcript

Use the fundamental theorem of calculus to find the derivative of the function 𝑦 is equal to the integral from four to five 𝑥 plus three of two 𝑡 divided by two plus 𝑡 to the fifth power with respect to 𝑡.

We need to find the derivative of a function 𝑦. And we might not be sure what to differentiate this with respect to because we see that there are two variables, 𝑡 and 𝑥. However, we need to notice that 𝑡 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 𝑦 is a function of 𝑥, and we need to differentiate 𝑦 with respect to 𝑥. And we’re told to do this by using the fundamental theorem of calculus.

So let’s start by recalling the fundamental theorem of calculus. This says if lowercase 𝑓 is a continuous function on a closed interval from 𝑎 to 𝑏 and capital 𝐹 of 𝑥 is the definite integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡, then we can find an expression for capital 𝑓 prime of 𝑥. That’s the derivative of the integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡 with respect to 𝑥. And it’s equal to lowercase 𝑓 of 𝑥 for all values of 𝑥 in the open interval from 𝑎 to 𝑏.

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

But we can get around this problem. Let’s start by setting 𝑢 equal to five 𝑥 plus three. So, in our definition for 𝑦, we’ll substitute 𝑢 is equal to five 𝑥 plus three into our upper limit. This means that 𝑦 is equal to the definite integral from four to 𝑢 of two 𝑡 divided by two plus 𝑡 to the fifth power. But remember, we’re still trying to differentiate 𝑦 with respect to 𝑥. We want to find an expression for the derivative of 𝑦 with respect to 𝑥, so we’ll differentiate both sides of this equation with respect to 𝑥.

This gives us 𝑦 prime of 𝑥 is equal to d by d𝑥 of the integral from four to 𝑢 of two 𝑡 divided by two plus 𝑡 to the fifth power with respect to 𝑡. And using exactly the same argument we used at the start of this video, we can see that this integral is a function in 𝑢. So we’re trying to differentiate a function of 𝑢 with respect to 𝑥. But remember, 𝑢 is a function in 𝑥, so we can do this by using the chain rule. So let’s recall what the chain rule tells us.

The chain rule tells us if 𝑦 is a function in 𝑢 and 𝑢 is a function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. And this is exactly the situation we have here. 𝑦 is a function in 𝑢 and 𝑢 is a function in 𝑥. So we can find an expression for d𝑦 by d𝑥 by finding d𝑦 by d𝑢 and multiplying this by d𝑢 by d𝑥. So by using the chain rule, we have that 𝑦 prime of 𝑥 is equal to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥.

And now, we can just apply the fundamental theorem of calculus to evaluate this first derivative. We’re differentiating our integral with respect to 𝑢. The lower limit of integration is a constant, and the upper limit of integration is 𝑢, and our integrand is a rational function. This means it’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 𝑡 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 𝑢.

So by using the fundamental theorem of calculus, we have d𝑦 by d𝑢 is equal to two 𝑢 divided by two plus 𝑢 to the fifth power. But remember, we still need to multiply this by d𝑢 by d𝑥. And remember, 𝑢 is a function in 𝑥. It’s five 𝑥 plus three. This is a linear function, so its derivative with respect to 𝑥 will be the coefficient of 𝑥, which is five. So we have now shown that 𝑦 prime of 𝑥 will be equal to two 𝑢 divided by two plus 𝑢 to the fifth power multiplied by five.

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