Video Transcript
Use the fundamental theorem of calculus to find the derivative of the function ℎ of 𝑥 is equal to the definite integral from two to 𝑒 to the power of five 𝑥 of negative the natural logarithm of 𝑡 with respect to 𝑡.
In this question, we’re given a function ℎ of 𝑥, which is defined as a definite integral where the upper limit of integration is a function in 𝑥. And we need to determine the derivative of this function. We’re told to do this by using the fundamental theorem of calculus.
So let’s start by recalling the fundamental theorem of calculus. In fact, we’ll only recall part of this. It 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 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 gives us a way of differentiating an integral where the upper limit of integration is 𝑥.
However, in our case, our upper limit of integration is not 𝑥. In fact, it’s a function in 𝑥, 𝑒 to the power of five 𝑥. One way of getting around this problem would be to use the substitution 𝑢 of 𝑥 is equal to 𝑒 to the power of five 𝑥. We could then find ℎ prime of 𝑥 by using a combination of the fundamental theorem of calculus and the chain rule, and this would work. However, we can also just do this in the general case to rewrite our fundamental theorem of calculus.
We have 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 𝑢 of 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡 and if 𝑢 of 𝑥 is a differentiable function, then by using our standard statement of the fundamental theorem of calculus and the chain rule, we get that capital 𝐹 prime of 𝑥 is equal to lowercase 𝑓 evaluated at 𝑢 of 𝑥 times the derivative of 𝑢 of 𝑥 with respect to 𝑥 as long as 𝑢 of 𝑥 is in the open interval from 𝑎 to 𝑏.
And now we can just apply this version to our function ℎ of 𝑥. First, 𝑢 of 𝑥 is 𝑒 to the power of five 𝑥 is a differentiable function because it’s an exponential function. Next, capital 𝐹 of 𝑥 is the function we’re differentiating. In this case, that’s ℎ of 𝑥. Remember that lowercase 𝑓 of 𝑡 will be our integrand. In this case, that’s negative the natural logarithm of 𝑡. Finally, the lower limit of our integral will be 𝑎. In this case, 𝑎 is equal to two.
And there’s one more thing we need to do. Remember, to use the fundamental theorem of calculus, we always need to check where our integrand is continuous. In our case, our integrand is negative the natural logarithm of 𝑡. And we know the natural logarithm function is continuous across its entire domain. So lowercase 𝑓 of 𝑡 will be continuous for all values of 𝑡 greater than zero.
Remember, our value of 𝑎 is equal to two. So in this case, we can see as long as 𝑏 is bigger than two, our function lowercase 𝑓 will be continuous on the closed interval from two to 𝑏. This means we can now use our result for the modified version of the fundamental theorem of calculus. This gives us that ℎ prime of 𝑥 will be equal to lowercase 𝑓 evaluated at 𝑢 of 𝑥 times the derivative of 𝑢 of 𝑥 with respect to 𝑥.
And now we can just start evaluating this expression. We know 𝑢 of 𝑥 is 𝑒 to the power of five 𝑥 and lowercase 𝑓 of 𝑥 is negative the natural logarithm of 𝑡. So this gives us ℎ prime of 𝑥 is equal to negative the natural logarithm of 𝑒 to the power of five 𝑥 multiplied by the derivative of 𝑒 to the power of five 𝑥 with respect to 𝑥. And now we can just start evaluating this expression.
First, in our first term, remember, the natural logarithm function and the exponential function are inverses. So the natural logarithm of 𝑒 to the power of five 𝑥 is just equal to five 𝑥. So we can simplify the first part of this expression to give us negative five 𝑥.
Now, all we need to evaluate is this derivative. And we can do this by recalling one of our standard derivative results for exponential functions. For any real constant 𝑎, the derivative of 𝑒 to the power of 𝑎𝑥 with respect to 𝑥 is equal to 𝑎 times 𝑒 to the power of 𝑎𝑥. In our case, the value of 𝑎 is five. So differentiating this, we get five times 𝑒 to the power of five 𝑥. Then we can just rearrange this to get negative 25𝑥𝑒 to the power of five 𝑥. And this is our final answer.
Therefore, by using the fundamental theorem of calculus and the chain rule on our function ℎ of 𝑥, we were able to show that ℎ prime of 𝑥 is equal to negative 25𝑥 times 𝑒 to the power of five 𝑥.
