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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.
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Here, we have a function π¦ of π₯, which is defined by an integral.
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To find this derivative, instead of evaluating the integral directly, weβre gonna be using the fundamental theorem of calculus.
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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 πΉ of π₯, which is defined by the integral between π and π₯ of lowercase π of π‘ with respect to π‘.
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Then πΉ prime of π₯ is equal to lowercase π of π₯ for all π₯ on the open interval between π and π.
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Letβs now check that we have the correct form to use.
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The function defined by the integral is π¦ of π₯ instead of capital πΉ of π₯.
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Our integrand is a function which is in fact continuous over the entire set of real numbers.
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And this is lowercase π of π instead of lowercase π of π‘.
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The lower limit of our integral is two, which is a constant.
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However, here we run into the problem and that the upper limit is not π₯ but rather is π₯ to the power of four.
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And this is a function of π₯.
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To use the fundamental theorem of calculus, weβll therefore need to come up with a modification.
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First, we express π¦ of π₯ in a slightly tidier form.
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Next, weβll define our troublesome upper limit as something else, for example, the variable π’.
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We then have that π¦ of π₯ is equal to the integral between two and π’ of π of π dπ.
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Now, we need to find π¦ prime of π₯, which is π¦ of π₯ differentiated with respect to π₯.
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We can write this as d by dπ₯ of the integral which we formed here.
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This is the step at which the problem would occur.
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We canβt directly use the fundamental theorem of calculus to evaluate, yet, since our upper limit does not match the variable π₯.
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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₯.
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Of course, here we have a dπ¦ by dπ₯.
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At this crucial step, the chain rule allows us to re express this as dπ¦ by dπ’ times dπ’ by dπ₯.
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Looking at this part of the equation, we now see that it can be evaluated using the fundamental theorem of calculus since weβre taking a derivative with respect to π’ and our upper limit, is indeed π’.
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You may be able to see this more clearly by observing that the form now matches as shown here.
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Using the theorem weβre able to say that this is equal to π of π’.
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And hence, π¦ Prime of π₯ is equal to π of π’ dπ’ by dπ₯.
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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.
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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.
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We now recall that π of π is equal to five cos squared five π.
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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.
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Next, we need to differentiate π₯ the power of four with respect to π₯.
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And of course, this is four π₯ cubed.
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Multiplying these two things together, we are left with 20π₯ cubed times cos squared five π₯ to the power of four.
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And finally, we have reached the answer to our question, since this is π¦ prime of π₯.