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.
 A: Recall that if $f$ is invertible, then it is bijective. So, there exists a unique $y$ such that $y = f(x)$ on the domain of $g$, which is seen to be $(f(a),f(b))$ since we have an increasing function. By definition of the derivative:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(z) - g(f(x))}{z-f(x)} = \lim_{z \rightarrow f(x)} \frac{g(z) - x}{z-f(x)}$$
Now, as we tend $z$ closer and closer to $f(x)$, eventually it will have to belong to $(f(a),f(b))$, which means we can find another $x_z$ such that $f(x_z) = z$. We now choose $z$ sufficiently close, and take advantage of this fact. We then have: 
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(f(x_z)) - x}{f(x_z)-f(x)} = \lim_{z \rightarrow f(x)} \frac{x_z - x}{f(x_z)-f(x)}$$
We see that this final limit tends to $\frac{1}{f'(x)}$.
A: We know that $g$ is the inverse of $f$ defined on $(f(a),f(b))$. This means that $g\circ f:(a,b)\rightarrow (a,b)$ is the identity function, that is
$$g(f(x)) = x.$$
This function is clearly differentiable with derivative 
$$\frac{d}{dx}(g\circ f)(x) = 1$$
for all $x\in (a,b)$. This is the same as saying that
$$\lim_{t\rightarrow x}\frac{g(f(t))-g(f(x))}{t-x}=1.$$
Now let $s,y\in (f(a),f(b))$ and consider the difference quotient
$$\frac{g(s)-g(y)}{s-y}.$$
We know that there exists unique $t,x\in (a,b)$ such that $f(t) = s$ and $f(x) = y$ but then
$$\frac{g(s)-g(y)}{s-y} = \frac{g(f(t))-g(f(x))}{t-x}\cdot \frac{t-x}{f(t)-f(x)} =\frac{g(f(t))-g(f(x))}{t-x}\cdot \frac{1}{\frac{f(t)-f(x)}{t-x}}\rightarrow 1\cdot \frac{1}{f'(x)}$$
as $s\rightarrow y$. 
Note that we can apply the change of variables without knowing anything of $g$ for denote 
$$h(s) = \frac{g(s)-g(y)}{s-y}$$
then we know that $h(f(t))\rightarrow \frac{1}{f'(x)}$ as $t\rightarrow x$ (assuming that $y = f(x)$). Let $\varepsilon$ be given then we can find a $\delta>0$ such that $t\in (x-\delta,x+\delta)$ implies that $|h(f(t))-\frac{1}{f'(x)}|<\varepsilon$ but then if $s\in (f(x-\delta),f(x+\delta))$ then $|h(s)-\frac{1}{f'(x)}|<\varepsilon$.
A: Lemma: $g:U \rightarrow \mathbb{R}$ is differentiable at $x$ with derivative $A$ iff there exists a function $\phi:U \rightarrow \mathbb{R}$ continuous at $x$ such that $g(t)-g(x)=\phi(t)(t-x)$ for all $t \in U$ and $\phi(x)=A$.
The above lemma should be easy to prove. Apply it. 
