### Video Transcript

Use the fundamental theorem of
calculus to find the derivative of the function π¦ of π₯ equals the integral between
two and π₯ to the power of four of five cos squared five π with respect to π.

Here, we have a function π¦ of π₯,
which is defined by an integral. To find this derivative, instead of
evaluating the integral directly, weβre gonna be using the fundamental theorem of
calculus. The first part of the theorem tells
us that if lowercase π is a continuous function on the closed interval between π
and π and we have another function capital πΉ of π₯, which is defined by the
integral between π and π₯ of lowercase π of π‘ with respect to π‘. Then πΉ prime of π₯ is equal to
lowercase π of π₯ for all π₯ on the open interval between π and π. Letβs now check that we have the
correct form to use.

The function defined by the
integral is π¦ of π₯ instead of capital πΉ of π₯. Our integrand is a function which
is in fact continuous over the entire set of real numbers. And this is lowercase π of π
instead of lowercase π of π‘. The lower limit of our integral is
two, which is a constant. However, here we run into the
problem and that the upper limit is not π₯ but rather is π₯ to the power of
four. And this is a function of π₯. To use the fundamental theorem of
calculus, weβll therefore need to come up with a modification. First, we express π¦ of π₯ in a
slightly tidier form. Next, weβll define our troublesome
upper limit as something else, for example, the variable π’. We then have that π¦ of π₯ is equal
to the integral between two and π’ of π of π dπ.

Now, we need to find π¦ prime of
π₯, which is π¦ of π₯ differentiated with respect to π₯. We can write this as d by dπ₯ of
the integral which we formed here. This is the step at which the
problem would occur. We canβt directly use the
fundamental theorem of calculus to evaluate, yet, since our upper limit does not
match the variable π₯. One tool that we can use to move
forward is the chain rule which tells us that dπ¦ by dπ₯ is equal to dπ¦ by dπ’
times dπ’ by dπ₯. Of course, here we have a dπ¦ by
dπ₯. At this crucial step, the chain
rule allows us to re express this as dπ¦ by dπ’ times dπ’ by dπ₯. Looking at this part of the
equation, we now see that it can be evaluated using the fundamental theorem of
calculus since weβre taking a derivative with respect to π’ and our upper limit, is
indeed π’.

You may be able to see this more
clearly by observing that the form now matches as shown here. Using the theorem weβre able to say
that this is equal to π of π’. And hence, π¦ Prime of π₯ is equal
to π of π’ dπ’ by dπ₯. In fact, this is an extremely
useful generalization that we can use when the limits of our integral involve a
function of π₯ rather than π₯ itself. To be forward with our question, we
can now substitute back into the equation using our definition, which is that π’ is
equal to π₯ to the power of four. We now recall that π of π is
equal to five cos squared five π. Replacing our π with π₯ to the
power of four, we get that π of π₯ to the power of four is equal to five cos
squared five π₯ to the power of four.

Next, we need to differentiate π₯
the power of four with respect to π₯. And of course, this is four π₯
cubed. Multiplying these two things
together, we are left with 20π₯ cubed times cos squared five π₯ to the power of
four. And finally, we have reached the
answer to our question, since this is π¦ prime of π₯.