How to prove a weak version of Inverse Function Theorem using Extreme Value Theorem?

Question

Suppose that $f:(a,b) \rightarrow \mathbb{R}$ is continuously differentiable with $f'(x)>0$ for all $x \in (a,b)$. Let $g$ be the inverse function of $f$. How can I prove that $g$ is differentiable on its domain and that $g'(f(x))=\frac{1}{f'(x)}$ for all $x \in (a,b)$? (Probably using the Extreme Value Theorem?)

Answer 1

If $g$ denotes the inverse of $f,$ then we have that $$\lim_{y\to y_0}\frac{g(y)-g(y_0)}{y-y_0}=\lim_{x\to x_0}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}=\lim_{x\to x_0}\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)}.$$ Thus we get that $g$ is differentiable and $g'(f(x))=\frac{1}{f'(x)}.$