Help complete this proof of the inverse function derivative theorem

Question

Let $f$ be a differentiable injection on $[a,b]$. Show $(f^{-1})'(x) = 1/f'(f^{-1}(x))$.

Note: Proofs are available. This question asks for help completing this proof.

Partial Proof: For arbitrary $x$ in the domain of $f^{-1}$, write

$$\begin{align} \lim_{h \to 0}\frac {f^{-1}(x+h) - f^{-1}(x)}{h} &= \lim_{h \to 0}\frac {f^{-1}(x+h) - f^{-1}(x)}{f(f^{-1}(x+h)) - f(f^{-1}(x))}\\ &= \left[\lim_{h \to 0}\frac {f(f^{-1}(x+h)) - f(f^{-1}(x))}{f^{-1}(x+h) - f^{-1}(x)} \right]^{-1} \tag{1} \\ &= \left [f'(f^{-1}(x)) \right ]^{-1}. \tag{2} \end{align}$$

Re (1): Since $a \mapsto a^{-1}, a \neq 0$ is continuous, and $\lim f(g(x)) = f(\lim g(x))$ for continuous $f$.

Re (2): I'm having trouble justifying this step. It is clearly true that as $h \to 0$, then $f^{-1}(x+h) \to f^{-1}(x)$. It is likewise true that we can interchange limits for invertible (or continuous) functions, both of which are satisfied here, and use this to change the variable of limitation. I believe these two facts together justify Line (2), but am struggling to turn this into a clear and direct argument.

Questions:

1. How do I justify Line (2)?
2. Is there an alternate approach that is simpler and avoids the complexity here?
3. Is the rest of the proof correct?

I prefer responses that point me in the right direction and help me draw the correct conclusion.

Answer 1

It may go without saying (I'm not sure about your needs) that $f^{-1}$ exists because $f$ is injective.

Questions:

1. Since $f^{-1}$ is continuous, sending $h\to 0$ also sends $f^{-1}(x+h)-f^{-1}(x)\to 0$. Then, you can rewrite the numerator in terms of a new $h$. Alternatively, use the derivative definition

$$f^\prime(x)=\lim_{y\to x} \frac{f(y)-f(x)}{y-x}$$

2. There's a graphical approach that is very nice. Try thinking about inverses as reflecting about $y=x$.

3. You never explicitly used injectivity or differentiability. Do you need to? For (1), you could add a reason for why $a\neq 0$.

Answer 2

The theorem is false. What if $f(x)=x^3$ on $[-1,1]$ ? Certainly $f$ is differentiable and injective (in fact bijective), but its inverse is not differentiable at $0.$

The theorem should state something like this: If $f$ is continuously differentiable with $f'(x) \ne 0,$ then $f$ is locally invertible around $x$ and $(f^{-1})'(f(x))=\frac{1}{f'(x)}.$ This is a very special case of the Inverse Function Theorem.