Let us first review the first part of the Fundamental Theorem, I will use the formulation used in Analysis I by Terence Tao:
Let a< b be real numbers, and let f:[a,b]toRR be a Riemann integrable function. Let F:[a,b]toRR be the function
F(x)=int_a^xf(t)dt.
Then F is continuous. Furthermore, if x_0in[a,b] and f continuous in x_0, then F is differentiable in x_0, and F'(x_0)=f(x_0).
So now we want to know the derivative of
F(x)=int_(e^x)^5 5sin^5(t)dt.
We note that if we define G(x)=int_(x)^5 5sin^5(t)dt, then F(x)=G(e^x). So in order to find the derivative of F, we use the chain rule, so:
F'(x)=G'(e^x)d/dxe^x=G'(e^x)e^x.
Now we need to know the derivative of G. We note that in the theorem, the variable x serves as an upper limit of integration, while in G it serves as a lower limit. This changes the way we can find the derivative slightly, as I will show next.
Suppose you have such a function f as given in the theorem, then let H,G:[a,b]toRR such that:
H(x)=int_a^xf(t)dt and G(x)=int_x^bf(t)dt.
Now note that (H+G)(x)=H(x)+G(x)=int_a^bf(t)dt.
Suppose f continuous in x_0in[a,b], then H'(x_0)=f(x_0), using the fundamental theorem. Furthermore (H+G)(x) is constant, so (H+G)'(x)=0. Since G(x)=(H+G)(x)-H(x):
G'(x_0)=(H+G)'(x_0)-H'(x_0)=0-f(x_0)=-f(x_0).
So G'(x_0)=-f(x_0).
Applying this to G(x)=int_(x)^5 5sin^5(t)dt, we note that 5sin^5(t) is continuous.
Therefore if x<=5, it serves as a lower limit, so G'(x)=-5sin^5(x).
If x>5, we use the fact that int_(x)^5 5sin^5(t)dt=-int_5^x 5sin^5(t)dt. Since x now serves as an upper limit, we may use the original theorem and once again see G'(x)=-5sin^5(x).
We already saw F'(x)=G'(e^x)e^x, so F'(x)=-5sin^5(e^x)e^x.
