Prove the inverse of this differentiable function is differentiable? Suppose we have a differentiable function $ g $ that maps from a real interval $ I $ to the real numbers and suppose $ g'(r)>0$ for all $ r$ in $ I $. 
Then I want to show that $ g^{-1}$ is differentiable on $g(I). $
Intuitively this makes sense but I can't come up with a neat proof. I was thinking to use the mean value theorem but I'm not sure if that would get me anywhere.

By definition of derivative we have
$$g'(r)=\lim_{x\to r}\frac{g(x)-g(r)}{x-r}=\lim_{x\to r}\frac{g(x)-g(r)}{g^{-1}(g(x))-g^{-1}(g(r))}$$
Since $g'(r)>0$, this means 
$$\lim_{x\to r}\frac{g^{-1}(g(x))-g^{-1}(g(r))}{g(x)-g(r)}=\frac{1}{g'(r)}$$
Let $y\in g(I)$, say $g(x)=y$ for some $x\in I$, and let $y_n\to y$. Then for each $n$ there is a unique $x_n\in I$ with $g(x_n)=y_n$. Further, one can show that if $f$ is continuous and strictly monotone on an interval $I$, then $f^{-1}$ is continuous on $f(I)$, hence we have $x_n\to x$, it follows that
$$\lim_{n\to\infty}\frac{g^{-1}(y_n)-g^{-1}(y)}{y_n-y}=\lim_{n\to\infty}\frac{x_n-x}{g(x_n)-g(x)}=\frac{1}{g'(x)}$$
Hence $g^{-1}$ is differentiable.
