real analysis - Prove the inverse of a strictly increasing function is differentiable.

So, I was given the following problem as part of a homework assignment.

Suppose $f'(x) > 0$ in $(a,b)$. Prove that $f$ is strictly increasing in $(a,b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that
$$g'(f(x)) = \frac{1}{f'(x)}$$

I have proven that $f$ is strictly increasing in $(a,b)$, and I could prove that $g'(f(x)) = 1/f'(x)$ if I could prove that $g$ is differentiable. The problem is that I am having trouble with a proof of that. Any advice?

Also, as a reference, this is exercise 5.2 from Baby Rudin.