I have $\displaystyle f(x)=\int_0^x 1+sin(sin(t))dt$ and I must find $\displaystyle [f^{-1}(0)]'$. I'm asking you to help me finding the inverse function of f, I'll try to do the derivative.

One of my guesses would be that as I have an integral of a function, the inverse operation is the derivative of the integral. In other words I would say that maybe the inverse function of $\displaystyle f$ is $\displaystyle 1+sin(sin(x))$. But looking back, if I replace this in the upper limit of the integral of $\displaystyle f(x)$, I doubt it will give me the identity. I also know that $\displaystyle 1+sin(sin(x))$ is $\displaystyle f'(x)$. Hmm... can you help me a bit?