suppose , find f'(Pi^(1/3))
Im not sure weather i integrate or diffrenciate.
This is an application of the chain rule. Say that $\displaystyle F(y)=\int\sin(5y)\,dy$. Thus, we are given that $\displaystyle f(t)=F(-t^3)-F(3)$.
Now take the derivative. (Remember, $\displaystyle F(3)$ is a constant, so its derivative is $\displaystyle 0$.)
$\displaystyle f'(t)=\underbrace{F'(-t^3)=\sin(5\cdot-t^3)\cdot\frac{d}{dt}[-t^3]}_{chain~rule}=\sin(-5t^3)\cdot-3t^2=\boxed{3t^2\sin(5t^3)}$
(Recall that sine is an odd function so $\displaystyle \sin(-5t^3)=-\sin(5t^3)$.)