Video: Finding the Derivative of a Function Defined by an Integral Where One of the Integral’s Limits Is a Power Function

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑦(π‘₯) = ∫_(2) ^(π‘₯⁴) 5 cosΒ² (5πœƒ) dπœƒ.

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Video Transcript

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

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

The function defined by the integral is 𝑦 of π‘₯ instead of capital 𝐹 of π‘₯. Our integrand is a function which is in fact continuous over the entire set of real numbers. And this is lowercase 𝑓 of πœƒ instead of lowercase 𝑓 of 𝑑. 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, we’ll therefore need to come up with a modification. First, we express 𝑦 of π‘₯ in a slightly tidier form. Next, we’ll define our troublesome upper limit as something else, for example, the variable 𝑒. We then have that 𝑦 of π‘₯ is equal to the integral between two and 𝑒 of 𝑓 of πœƒ dπœƒ.

Now, we need to find 𝑦 prime of π‘₯, which is 𝑦 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 can’t 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 d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. Of course, here we have a d𝑦 by dπ‘₯. At this crucial step, the chain rule allows us to re express this as d𝑦 by d𝑒 times d𝑒 by dπ‘₯. Looking at this part of the equation, we now see that it can be evaluated using the fundamental theorem of calculus since we’re taking a derivative with respect to 𝑒 and our upper limit, is indeed 𝑒.

You may be able to see this more clearly by observing that the form now matches as shown here. Using the theorem we’re able to say that this is equal to 𝑓 of 𝑒. And hence, 𝑦 Prime of π‘₯ is equal to 𝑓 of 𝑒 d𝑒 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 𝑒 is equal to π‘₯ to the power of four. We now recall that 𝑓 of πœƒ is equal to five cos squared five πœƒ. Replacing our πœƒ with π‘₯ to the power of four, we get that 𝑓 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 𝑦 prime of π‘₯.

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