### Video Transcript

Find the derivative of the function

𝑔 of 𝑥 is equal to the definite integral from three 𝑥 to four 𝑥 of 𝑢 squared

minus three all divided by 𝑢 squared plus five with respect to 𝑢.

We’re given a function 𝑔 of

𝑥. And we can see this is a definite

integral where both of the limits of integration are functions in 𝑥. And we’re asked to find the

derivative of this function. Since we’re trying to differentiate

a function where the limits of integration are functions in 𝑥, we’ll try and do

this by using the fundamental theorem of calculus.

So we’ll start by recalling the

following part of the fundamental theorem of calculus. If lowercase 𝑓 is a continuous

function on a closed interval from 𝑎 to 𝑏 and we have capital 𝐹 of 𝑥 is the

definite integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑢 with respect to 𝑢, then

capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 of 𝑥 for all values of 𝑥 in

the open interval from 𝑎 to 𝑏. In other words, this part of the

fundamental theorem of calculus gives us a way of differentiating a function defined

by an integral.

However, the function we’re asked

to differentiate in the question, 𝑔 of 𝑥, is very different to our definition of

capital 𝐹 of 𝑥. First, the lower limit of

integration is supposed to be a constant. However, we’re given this as a

function in 𝑥. There’s a similar story for our

upper limit of integration. It’s supposed to be 𝑥, but we’re

given a function in 𝑥. However, we can get around both of

these problems by using what we know about definite integrals and the chain

rule.

The first thing we’re going to need

to recall is one of our rules for definite integrals. To get around the problem of not

having a constant limit of integration, we’re going to want to split our definite

integral into two definite integrals. We know the following rule for

definite integrals. The integral from 𝑎 to 𝑏 of 𝑓 of

𝑢 with respect to 𝑢 is equal to the integral from 𝑐 to 𝑏 of 𝑓 of 𝑢 with

respect to 𝑢 plus the integral from 𝑎 to 𝑐 of 𝑓 of 𝑢 with respect to 𝑢. But there’s some important

information about this rule which we need to clarify.

The most important of these is this

will only be true if our function 𝑓 is integrable on all of these domains of

integration. In particular, our function 𝑓 must

be integrable on the closed interval from 𝑏 to 𝑐 and the closed interval from 𝑐

to 𝑎. We want to use this to rewrite our

function 𝑔 of 𝑥. So by using the upper limit of

integration as four 𝑥, the lower limit of integration as three 𝑥, and our function

lowercase 𝑓 of 𝑢 as 𝑢 squared minus three divided by 𝑢 squared plus five, we

would get the integral from 𝑐 to four 𝑥 of 𝑢 squared minus three divided by 𝑢

squared plus five with respect to 𝑢 plus the integral from three 𝑥 to 𝑐 of 𝑢

squared minus three all divided by 𝑢 squared plus five with respect to 𝑢.

But remember, we also need our

function to be integrable on both of these domains of integration. In particular, we need to choose a

value of the constant 𝑐. In this case, we could actually

choose any value of the constant 𝑐. To explain why, let’s take a look

at our integrand. Our integrand is the quotient of

two polynomials. This means it’s a rational

function. And we know rational functions will

be continuous across their entire domain. And if a function is continuous on

an interval, then it’s also integrable on that interval. So all we need to do is check the

domain of our function lowercase 𝑓 of 𝑢. It’s a rational function, so it

will be defined for all values of 𝑢 except where its denominator is equal to

zero.

And if we solve the denominator 𝑢

squared plus five is equal to zero, we’ll see that we get no real solutions. So our integrand is defined for all

real values of 𝑢. Meaning it’s continuous for all

real values of 𝑢, meaning it’s integrable on any interval. So in this case, we can actually

pick any value of 𝑐. However, this won’t always be the

case. In this case, we’ll choose 𝑐 is

equal to zero. We’ve now split this into two

definite integrals, each of which is closer to being able to use the fundamental

theorem of calculus.

There’s a little bit more

manipulation we need to do before we can use the fundamental theorem of

calculus. In our second integral, we want to

swap the order of our limits of integration. Remember, we can do this by

multiplying the entire integral by negative one. Doing this, we’ve rewritten our

second integral as negative one times the definite integral from zero to three 𝑥 of

𝑢 squared minus three all divided by 𝑢 squared plus five with respect to 𝑢.

And now, we’re almost ready to use

the fundamental theorem of calculus to evaluate these definite integrals. The only problem is our upper limit

of integration is a function in 𝑥 instead of 𝑥. And there are two ways of getting

around this problem. First, we could use the

substitution 𝑣 of 𝑥 is equal to four 𝑥. Then we can differentiate our first

integral by using the chain rule. We could do the same with our

second integral, setting 𝑣 of 𝑥 to be three 𝑥 and then evaluating its derivative

by using the chain rule. And this would work. However, we can also do this in the

general case with the fundamental theorem of calculus.

We have if 𝑣 of 𝑥 is a

differentiable function, then by using the chain rule, we get if lowercase 𝑓 is a

continuous function on the closed interval from 𝑎 to 𝑏 and capital 𝐹 of 𝑥 is the

definite integral from 𝑎 to 𝑣 of 𝑥 of lowercase 𝑓 of 𝑢 with respect to 𝑢, then

capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 evaluated at 𝑣 of 𝑥 times d𝑣

by d𝑥 as long as 𝑣 of 𝑥 is in the open interval from 𝑎 to 𝑏. So we were able to rewrite our

version of the fundamental theorem of calculus so long as 𝑣 of 𝑥 is a

differentiable function just by using the chain rule.

And now, we can see both of our

definite integrals are in this form. We’ll set 𝑣 one of 𝑥 to be four

𝑥 and 𝑣 two of 𝑥 to be three 𝑥. Usually, we would set our value of

𝑎 equal to zero in both cases and then check the intervals on which our integrand

is continuous. This is because to use the

fundamental theorem of calculus, we do need to check where our integrand is

continuous. However, we’ve already shown that

our integrand is continuous for all real values. Therefore, it will be continuous on

any closed interval.

We’re now ready to find an

expression for 𝑔 prime of 𝑥. We’ll differentiate each of our

integrals separately. To differentiate our first integral

by using the fundamental theorem of calculus, we do need to check that 𝑣 one of 𝑥

is differentiable. And it’s a linear function, so we

know it is differentiable. This means we can use this, and we

get 𝑓 evaluated at 𝑣 one of 𝑥 times d𝑣 one by d𝑥. And we get the same story in our

second integral. This time, 𝑣 two of 𝑥 will be

three 𝑥, which is also a linear function, so it’s differentiable. This means we need to subtract 𝑓

evaluated at 𝑣 two of 𝑥 times d𝑣 two by d𝑥.

So now, we found an expression for

𝑔 prime of 𝑥. And we can actually evaluate all of

these expressions. Let’s start by substituting in our

expressions for 𝑣 one of 𝑥 and 𝑣 two of 𝑥. This gives us 𝑔 prime of 𝑥 is

equal to lowercase 𝑓 evaluated at four 𝑥 times the derivative of four 𝑥 with

respect to 𝑥 minus lowercase 𝑓 evaluated at three 𝑥 multiplied by the derivative

of three 𝑥 with respect to 𝑥. And remember, from the fundamental

theorem of calculus, lowercase 𝑓 will be our integrand. So we’re now ready to find an

expression for 𝑔 prime of 𝑥.

First, to find 𝑓 evaluated at four

𝑥, we substitute 𝑢 is equal to four 𝑥 into our integrand, giving us four 𝑥 all

squared minus three all divided by four 𝑥 all squared plus five. Next, we need to multiply this by

the derivative of four 𝑥 with respect to 𝑥. Of course, this is a linear

function, so we know the derivative of this with respect to 𝑥 will be the

coefficient of 𝑥, which is four. And we can do exactly the same to

find an expression for our second term.

To find 𝑓 evaluated at three 𝑥,

we substitute 𝑢 is equal to three 𝑥 into our integrand, giving us three 𝑥 all

squared minus three all divided by three 𝑥 all squared plus five. And of course, we need to multiply

this by the derivative of three 𝑥 with respect to 𝑥, which we know is three. Finally, we’ll simplify this

expression by evaluating all of our exponents and multiplying through by our

coefficients. And by doing this, we get our final

answer. 𝑔 prime of 𝑥 is equal to four

times 16𝑥 squared minus three all divided by 16𝑥 squared plus five minus three

times nine 𝑥 squared minus three all divided by nine 𝑥 squared plus five.