Question about the proof for the derivative of the inverse function

Question

I am trying to prove that the inverse function of a differentiable function is also differentiable.

Here is what I did: There exists $x$ and $x_0$ such that $f(x)=y$ and $f(x_0)=y_0$. Thus: $$\lim_{y \to \ y_0} \frac{f^{-1}(y) - f^{-1}(y_0)}{y - y_0} = \lim_{y \to \ y_0} \frac{f^{-1}(f(x)) - f^{-1}(f(x_0)}{f(x) - f(x_0)}\\ =\lim_{y \to \ y_0} \frac{x - x_0}{f(x) - f(x_0)}=\lim_{y \to \ y_0} \frac{1}{\frac{f(x) - f(x_0)}{x - x_0}}.$$

I know that it's equal to $\frac{1}{f'(x_0)}$ but I don't know how to explain why the fact that $y$ approaches $y_0$ implies that $x$ approaches $x_0$ so I will be able to use the definition of the derivative.

Answer 1

We have $x=f^{-1}(y)$, so when $y \to y_0,\, x \to f^{-1}(y_0)$.

Also, we have $x_0=f^{-1}(y_0)$ , so $x \to x_0$.

Answer 2

Here is one version of the rule for derivative of inverse functions.

Theorem: Let $f$ be a real valued function defined in a certain neighborhood $I$ of $x_0$ and $y_0=f(x_0)$. Also let $f$ be strictly monotone and continuous on $I$ with $f'(x_0)\neq 0$. Then there is a neighborhood $J$ of $y_0$ such that the inverse function $f^{-1}$ is defined, strictly monotone and continuous on $J$ and further derivative of $f^{-1}$ at $y_0$ equals $1/f'(x_0)$.

The hypotheses of continuity and monotone nature of $f$ are needed to get a well behaved inverse. And the result above tells that $f^{-1}$ is continuous on $J$ and in particular at $y_0$ hence as $y=f(x) $ tends to $y_0$ the variable $x=f^{-1}(y)$ tends to $f^{-1}(y_0)=x_0$ and your proof can be completed.

Most introductory calculus books do not state precisely the conditions under which a theorem is valid. So you may miss such detail regarding continuity while dealing with derivatives of inverse functions.