Inverse of an integral function $$f(x) =\int_0^x  \sin(t)\,dt.$$
I have to compute $(f^{-1})'(0)$ and I know it involves the formula for the derivative of an inverse function $1/f'(f^{-1}(x))$ and that $f' = \sin(x)$ but I do not know how to find $f^{-1}$ of an inverse function.
 A: You need to find $(f^{-1})'(0)$, and you know from the formula you mentioned that $$(f^{-1})'(0) = \frac{1}{f'(f^{-1}(0))}.$$  Now, given that $$f(x) =\int_0^x  \sin(t)\,dt,$$ you have that $f(0) = 0$, and so $f^{-1}(0) = 0$ as well.  That simplifies the question to finding $$(f^{-1})'(0) = \frac{1}{f'(0)}.$$  You have also said that $f'(x) = \sin(x)$, and since $\sin(0) = 0$ you have that $(f^{-1})'(0)$ is undefined, and in fact $(f^{-1})'(x) \to \infty$ as $x\to 0^{+}$.
Alternatively, you can compute directly that $$f(x) = \int_{0}^{x}\sin(t)\,dt = -\cos(t)\bigg|_{0}^{x} = 1 - \cos(x)$$ and so $$f^{-1}(x) = \arccos(1-x).$$  Differentiating, then, gives us $$(f^{-1})'(x) = \frac{1}{\sqrt{1-(1-x)^{2}}} = \frac{1}{\sqrt{2x-x^{2}}}.$$  Again, we have that $(f^{-1})'(0)$ is undefined, and $(f^{-1})'(x) \to \infty$ as $x\to 0^{+}$.
A: if you have:
$$y=f^{-1}(x)\Rightarrow x=f(y)$$
$$\frac{dx}{dy}=f'(y)$$
$$\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$$
so:
$$\left[f^{-1}\right]'=\frac{1}{f'\circ f^{-1}}$$
Just in case you were interested in where that formula comes from.

$$f(x)=\int_0^x\sin(t)dt=\left[-\cos(t)\right]_0^x=1-\cos(x)$$
$$f'(x)=\sin(x)$$
now to find to inverse function solve the following for $y$:
$$x=1-\cos(y)$$
$$\cos(y)=1-x$$
$$y=\arccos(1-x)=f^{-1}(x)$$
now combine the two and you are done :)
