I know $y=f(x)$ is differentiable and invertible.
His inverse function is $x=g(y)$ which is also differentiable.
I have to prove that $g'(y)= 1/(f'(x))$.
I tried first with an baby example with $y=x^2$ and it turned out that this holds.
But I have no idea on how to prove this. Please provide some help!