Differentiability of inverse function (Rudin 5.2) From chapter 5, Exercise 2, of Rudin's Principles, 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've already shown that $f$ is strictly increasing and that $g$ exists. My question boils down to verification of the following proof of the remainder.
Let $(c,d)=f(a,b)$. Put $t, x\in (a,b)$ and $y=f(t),z=f(x)\in(c,d)$. Let $\varepsilon>0$ be given. We wish to show that there is some $\eta>0$ such that
$$
\left|\frac{g(y)-g(z)}{y-z}-\frac{1}{f'(x)}\right|=\left|\frac{t-x}{f(t)-f(x)}-\frac{1}{f'(x)}\right|<\varepsilon
$$
whenever $0<|y-z|=|f(t)-f(x)|<\eta$. By differentiability of $f$ at $x$, there is some $\delta>0$ such that $\left|\frac{t-x}{f(t)-f(x)}-\frac{1}{f'(x)}\right|<\varepsilon$ whenever $0<|t-x|<\delta$. Since $f$ is differentiable at $x$, $f$ is continuous at $x$, and $f(t)\to f(x)$ if and only if $t\to x$. In other words, for each $\eta_1>0$, there is some $\eta_2>0$ such that $0<|t-x|<\eta_1$ whenever $0<|f(t)-f(x)|<\eta_2$ and vice versa. Hence, there is some $\eta>0$ such that $0<|f(t)-f(x)|<\eta$ implies $0<|t-x|<\delta$ implies the claim, and we are done.
 A: The problem with your proof in its current form is the moment where you invoke continuity. It does not garantee at all the "vice versa" part (which is not true for a function constant on a part of its definition domain). 
Consider some $t \in ]a,b[$, we have $f'(t)>0$. Choose some $\epsilon>0$ small enough such that $\epsilon > \frac{\epsilon^2}{f'(t)(1+\epsilon^2)}$ (which you can since this tends to $0$ as a function of $\epsilon$)
$f$ is derivable in $t$ so $\forall \epsilon_1 >0, \exists \eta_1 > 0, \forall x \in [t-\eta_1, t+\eta_1], f'(t)-\epsilon_1<\frac{f(x)-f(t)}{x-t}<f'(t)+\epsilon_1 $
Take $\epsilon_1=f'(t)\epsilon^2$ and invert, this gives you
$\frac{1}{f'(t)(1-\epsilon^2)} >\frac{x-t}{f(x)-f(t)}>\frac{1}{f'(t)(1+\epsilon^2)}$
Substract $\frac{1}{f'(t)}$, the rest follows
Doing this for a small $\epsilon$ is enough. Then, you can reformulate this to include $g$ before introducing $t$.
A: All fine.
I would be happy if you had stopped starting with "Since $f$ is differentiable...". What is important is that $y-z$ takes any value, no matter if $f(t)-f(x)$ does or not (it does but this changes nothing in your argument).
