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

# Question Video: Finding the Derivative of a Function Defined by an Integral Where One of the Integralβs Limits Is a Power Function Mathematics • Higher Education

Use the Fundamental Theorem of Calculus to find the derivative of the function π¦(π₯) = β«_(2) ^(π₯β΄) 5 cosΒ² (5π) dπ.

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### 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, weβre 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 πΉ of π₯, which is defined by the integral between π and π₯ of lowercase π of π‘ with respect to π‘. Then πΉ prime of π₯ is equal to lowercase π of π₯ for all π₯ on the open interval between π and π. Letβs now check that we have the correct form to use.

The function defined by the integral is π¦ of π₯ instead of capital πΉ 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, weβll therefore need to come up with a modification. First, we express π¦ of π₯ in a slightly tidier form. Next, weβll 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 canβ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 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 weβre 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 weβre 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₯.

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