I'm reviewing old calculus notes, and we are given the inverse function theorem, note that invertible means injective here, and $f^{-1}:= f^{-1}(f(x))=x, \forall x \in D(f)$.
Theorem. If $f$ is invertible on $(x_0-\delta, x_0+\delta)$ for some $\delta>0$, and $f$ is differentiable at $x_0$, then $f^{-1}$ is invertible at $f(x_0)$, and $(f^{-1})'(f(x_0))=\frac{1}{f'(f^{-1}(x_0))}$, because
\begin{align}
\lim_{x\to x_0}\frac{f^{-1}(x)-f^{-1}(x_0)}{x-x_0}
&=\lim_{x=f(y)\to f(y_0)=x_0}\frac{f^{-1}(f(y))-f^{-1}(f(y_0))}{f(y)-f(y_0))}\\
&=\lim_{y\to y_0}\frac{y-y_0}{f(y)-f(y_0)}
\end{align}
Questions:
- We used the first condition to be able to say that $x=f(y),x_0=f(y_0)$, but where did we use the second condition here?
- What is going on with the indices of the limit after the last equality sign?
Best Answer
I would write the proof like this: \begin{align} \lim_{y\to y_0}\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0} &=\lim_{y=f(x)\to y_0=f(x_0)}\frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\\ &=\lim_{x\to x_0}\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)} \end{align} We can make the last transition because $\lim_{f(x)\to f(x_0)}(x-x_0)=0$ since $f$ is continuous. I also think the statement of the theorem contains an error, it should be $(f^{-1})'(x_0)=\frac{1}{f'(f^{-1}(x_0))}$ or (thanks to @TobyBartels), $(f^{-1})'(f(x_0))=\frac{1}{f'(x_0)}$