# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y=int_("[e"^x",5]") 5(sin(t))^5 dt#?

If

Let us first review the first part of the Fundamental Theorem, I will use the formulation used in Analysis I by Terence Tao:

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To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function ( y = \int_{e^x}^{5} 5(\sin(t))^5 dt ), we first need to express the function as a definite integral of a function of ( x ).

Let ( F(x) = \int_{e^x}^{5} 5(\sin(t))^5 dt ).

According to Part 1 of the Fundamental Theorem of Calculus, if ( F(x) ) is defined as above, then ( F'(x) = f(x) ), where ( f(x) ) is the integrand function with the upper limit replaced by ( x ).

In this case, we have:

( f(x) = 5(\sin(x))^5 ).

Therefore, the derivative of the function ( y = \int_{e^x}^{5} 5(\sin(t))^5 dt ) with respect to ( x ) is:

( \frac{dy}{dx} = f(x) = 5(\sin(x))^5 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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