# Question Video: Finding the Derivative of a Function Defined by an Integral Where Its Limits Contain a Variable Mathematics • Higher Education

Find the derivative of the function 𝑔(𝑥) = ∫_(3𝑥)^(4𝑥) (𝑢² − 3)/(𝑢² + 5) d𝑢.

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### 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.