f(x) = integral from 0 to g(x) of ((t+1)/(t^5+1)^.5)dt where g(x) = integral from 0 to sinx of (1+cos(t^3))dt find f'(pi) help please!
Is the question $\displaystyle f(x) = \int_0^{g(x)} {\frac{{t + 1}}{{\sqrt {{t^5} + 1} }}dt} $ where $\displaystyle g(x) = \int_0^{\sin (x)} {\left( {1 + \cos ({t^3})} \right)dt} $, find $\displaystyle f'(\pi)~?$
If that is it, then the good news is that no integrals are needed.
$\displaystyle f'(x) = \frac{{g(x) + 1}}{{\sqrt {{g^5}(x) + 1} }}g'(x)$ where $\displaystyle g'(x) = \left( {1 + \cos \left( {{{\sin }^3}(x)} \right)} \right)\left( {\cos (x)} \right)$.
Now just evaluate that for $\displaystyle x=\pi~.$
I simply have no idea what your question means. Here is the basic idea.
If each of $\displaystyle h(x)~\&~g(x)$ is a differentiable function and $\displaystyle f(x) = \int_{g(x)}^{h(x)} {\phi (t)dt} $ then $\displaystyle f'(x) = \phi \left( {h(x)} \right)h'(x) - \phi \left( {g(x)} \right)g'(x)$.