Derivative of inverse function proof verification

Question

Can someone verify whether my attempt to prove this theorem is correct?

Notice that I use a generalized definition of derivative:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Let $p$ both a point and a limit point of $E$. Then we define the derivative of $f$ at $p$ as the limit $f'(p) = \lim_{x \to p, x \in E \setminus \{p\}} \frac{f(x)-f(p)}{x-p}$, provided the limit exists.

Theorem: Let $f: E \subseteq \mathbb{R} \to F \subseteq \mathbb{R}$ be an invertible function that is differentiable at $p \in E$. Suppose that $f^{-1}: F \to E$ is continuous at $f(p)$ and that $f'(p) \neq 0$. Then $f^{-1}$ is differentiable at $f(p)$, and we have

$$(f^{-1})'(f(p)) = \frac{1}{f'(p)}$$

Proof: Before proving the theorem, we have to check that differentiation at $f(p)$ makes sense: we have to show that $f(p)$ is a limit point of $Y$.

Let $\epsilon > 0$. Because $f$ is differentiable at $p$, $f$ is continuous at $p$ and it follows that $|f(p)-f(x)| < \epsilon$ whenever $x \in (p- \delta, p + \delta) \cap E \setminus \{p\}$ for some $\delta > 0$. Notice: $f'(p) \neq 0$ implies that $f$ is not constant on $(p- \delta, p + \delta) \cap E \setminus \{p\}$ (if it were constant, we would have $f'(p) = \lim_{x \to p} \frac{f(x)-f(p)}{x-p} = \lim_{x \to p, x \in E \cap (p- \delta, p + \delta)\setminus \{p\}}\frac{f(x)-f(p)}{x-p} = 0)$. Combining these facts, we deduce that $0 < |f(x)-f(p)| < \epsilon$ for some $x \in E$, and $f(p)$ is a limit point of $F = f(E)$.

Define $$F: E \to \mathbb{R}: x \mapsto \begin{cases} \frac{f(x)-f(p)}{x-p} \quad x \neq p \\ f'(p) \quad x = p\end{cases}$$

Clearly, $F$ is continuous at $p$.

The theorem now follows from the following easy calculation:

$$(f^{-1})'(f(p)) = \lim_{y \to f(p)} \frac{f^{-1}(y)- f^{-1}(f(p))}{y-f(p)}$$

$$= \lim_{y \to f(p)}\frac{1}{\frac{f(f^{-1}(y))-f(p)}{f^{-1}(y)- p}}$$

$$ = \lim_{y \to f(p)} \frac{1}{F(f^{-1}(y))}$$

$$= \frac{1}{F( \lim_{y \to f(p)} f^{-1}(y))} = \frac{1}{F(p)} = \frac{1}{f'(p)}$$

However, some equalities need some explanation. The second equality is justified by noticing that $f^{-1}(y) - p = f^{-1}(y) - f^{-1}(f(p))$ is zero only when $y = f(p)$, so there are no trouble with dividing by zero. The fourth equality uses the continuity of $F$ at $p$ and the fifth equality follows from the continuity of $f^{-1}$ at $f(p) \quad \square$

Answer 1

This looks fine to me with one exception. You don't want to claim that $f$ isn't constant on $(p-\delta,p+\delta)$. You want to claim that $f(x)\ne f(p)$ for all $x$ with $0<|x-p|<\delta$. (The proof for that is not quite the proof you've supplied.) You need this because you need to know that as $y\to f(p)$ you can make sense of the inverse function (and you never have a division by zero problem). In some sense, the statement of your theorem is flawed, as you're assuming existence of the inverse function. That should be part of what you prove here — namely, that $f$ is one-to-one on some interval around $p$.

Answer 2

Your work will be done faster if not make use of the First Principle of Derivative.

We know that $f^{-1}(f(x))=x$, differentiate it combining with Chain Derivative, we have $$\begin{align}(f^{-1}(f(x)))'&=1\\f^{-1}{'}(f(x))f'(x)&=1\\\therefore f^{-1}{'}(f(x))&=\dfrac1{f'(x)}\end{align}$$

I am not too used in rigorous proof in derivative (such like the first principle) but I think your proof is good enough.