### 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 weβre told to do this by using
the fundamental theorem of calculus.

So letβs 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 πΉ 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₯. Thatβs the derivative of the
integral from π to π₯ of lowercase π of π‘ with respect to π‘ with respect to
π₯. And itβs 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
πΉ 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. Letβs start by setting π’ equal to
five π₯ plus three. So, in our definition for π¦, weβll
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, weβre 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 weβll 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 weβre 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 letβs 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. Weβre 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 itβs 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₯. Itβs 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 weβre 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.