My Proof for the Inverse Function Theorem Here's what I want to prove:
Let $f: (a,b) \to \mathbb{R}$ be a continuous injection, let $y_0 \in f[(a,b)]$ and let $f^{-1}: f[(a,b)] \to (a,b)$ be the inverse. Suppose that $f$ is differentiable at $x_0 = f^{-1}(y_0)$ so that $f'(x_0) \neq 0$. Then, $f^{-1}$ is differentiable at $y_0$ and:
$$(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$$

Proof Attempt:
We know that $f[(a,b)]$ is an interval since $f$ is continuous and that $f^{-1}$ is a continuous injection. Without loss of generality, assume that $f$ is strictly increasing. Then, $f^{-1}$ is strictly increasing. Then, we are told that the following limit exists:
$$f'(x_0) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$
$$\implies f'(x_0) = \lim_{x \to f^{-1}(y_0)} \frac{f(x)-y_0}{x-f^{-1}(y_0)}$$
Let $y = f(x)$. Since $f: (a,b) \to f[(a,b)]$ is a bijection, $f^{-1}(y) = x$. So, we now have:
$$f'(x_0) = \lim_{x \to f^{-1}(y_0)} \frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)}$$
The above is equivalent to the following predicate:
$$\forall \epsilon > 0: \exists \delta > 0: x \in (f^{-1}(y_0)-\delta,f^{-1}(y_0)+\delta) \setminus \{f^{-1}(y_0)\} \implies \left| \frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)} - f'(x_0) \right| < \epsilon$$
Then, since $f$ is a strictly increasing continuous function, it follows that:
$$x \in ((f^{-1}(y_0)-\delta,f^{-1}(y_0)+\delta) \setminus \{f^{-1}(y_0)\}$$
$$\iff y = f(x) \in (f(f^{-1}(y_0)-\delta),f(f^{-1}(y_0)+\delta)) \setminus \{y_0 \}$$
which implies that $y \in (y_0-\delta',y_0+\delta')$ for some $\delta' > 0$. So, in fact, we have shown that the following statement holds:
$$\forall \epsilon > 0: \exists \delta' > 0: 0 < |y-y_0| < \delta' \implies \left| \frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)} - f'(x_0) \right| < \epsilon$$
However, that means that:
$$\lim_{y \to y_0} \frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)} = f'(x_0) \neq 0$$
which further implies that:
$$\frac{1}{f'(x_0)} = \frac{1}{\lim_{y \to y_0} \frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)}} = \lim_{y \to y_0} \frac{1}{\frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)}} = \lim_{y \to y_0} \frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0} = (f^{-1})'(y_0)$$
as was desired. $\Box$
Does the proof above work? If it doesn't, then why? How can I fix it?
 A: It might be easier to do the following:
Let $g = f^{-1}$, we have $g(f(x)) = x$.
Suppose $t_n \to t=f(x_0)$, then (for sufficiently large $n$) we can find $x_n$ such that $t_n = f(x_n)$. Futhermore, we have $x_n \to x_0$.
Then ${g(t_n)-g(t) \over t_n-t} = {x_n -x_0 \over f(x_n)-f(x_0)} \to {1 \over f'(x_0)}$.
A: Your proof works fine, in my opinion it just lacks two details:

*

*You could add some more information about why $f[(a,b)]$ is an interval and why $f$ must be a strictly monotonic function. Both of these facts are consequences of the Intermediate Value Theorem.


*You should be more careful in the passage with $\delta'$. Although the conclusion is true the argument was wrong. You say that "$y\in(f(f^{-1}(y_0)-\delta),f(f^{-1}(y_0)+\delta))\setminus \{y_0\}$ implies that $y\in (y_0-\delta', y_0 + \delta')$ for some $\delta'>0$", which is not wrong, but notice that, in the next passage, when you conclude that
$$\forall \epsilon > 0: \exists \delta' > 0: 0 < |y-y_0| < \delta' \implies \left| \frac{y-y_0}{f^{-1}(y)-f^{-1}(y_0)} - f'(x_0) \right| < \epsilon,
$$
you are actually using the reverse implication, which is that "$y\in (y_0-\delta', y_0 + \delta')\setminus \{y_0\}$ implies $y\in(f(f^{-1}(y_0)-\delta),f(f^{-1}(y_0)+\delta))\setminus \{y_0\}$", not the one you stated.
Of course this can be fixed by something like:
$$
x \in (f^{-1}(y_0)-\delta,f^{-1}(y_0)+\delta) \setminus \{f^{-1}(y_0)\} \iff y\in(f(f^{-1}(y_0)-\delta),f(f^{-1}(y_0)+\delta))\setminus \{y_0\},
$$
and since $f(f^{-1}(y_0)-\delta)< y_0<f(f^{-1}(y_0)+\delta)$ there exists $\delta'>0$ such that
$$
y\in (y_0-\delta', y_0 + \delta')\setminus \{y_0\}\ \ \Rightarrow\ \ y\in(f(f^{-1}(y_0)-\delta),f(f^{-1}(y_0)+\delta))\setminus \{y_0\}.
$$
