### 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.