In class the other day we proved the following theorem regarding the derivative of an invertible, differentiable function $f$:
Theorem
Assume that $f \colon D \to V$ is a bijection, $x_0\in D$, $y_0=f(x_0)$, $f'(x_0) \neq 0$ exists and that $f^{-1}$ is continuous at $y_0$. Then the derivative $(f^{-1})'(y_0)$ exists and $(f^{-1})'(y_0) = \frac{1}{f'(x_0)}.$
Proof
Consider the difference quotient \begin{equation*} \frac{f^{-1}(y_0+k)-f^{-1}(y_0)}{k}. \end{equation*}
Set $h = f^{-1}(y_0+k)-f^{-1}(y_0)$ so that $f^{-1}(y_0+k) = x_0+h$. We then have $f(f^{-1}(y_0+k)) = y_0+k = f(x_0+h)$.
From this we obtain
\begin{equation} k = f(x_0+h)-y_0 = f(x_0+h)-f(x_0) \end{equation}
and so
\begin{equation*} \frac{f^{-1}(y_0+k)-f^{-1}(y_0)}{k} = \frac{h}{f(x_0+h)-f(x_0)} \to \frac{1}{f'(x_0)} \text{ when } k\to0. \end{equation*}
What confuses me is why we can assume that $f^{-1}(y_0+k)$ exists for sufficiently small $k$, i.e. that $y_0$ is an inner point of $V$. I can't seem to wrap my head around why. I know that since we assume that $f$ is differentiable at $x_0$, we implicitly assume that $x_0$ is an interior point of $D$. I also know that since $f$ is differentiable at $x_0$, $f$ is also continuous at $x_0$. Does this somehow imply that $y_0 = f(x_0)$ is an interior point of $V$?
Suppose that $f'(x_0)>0$. Then, in some interval $(x_0-\delta,x_0+\delta)$, you have$$x<x_0\implies f(x)<f(x_0)=y_0\text{ and }x>x_0\implies f(x)>f(x_0)=y_0.$$Since $f\bigl((x_0-\delta,x_0+\delta)\bigr)$ is an interval (since $f$ is continuous), it follows that, yes, $y_0$ is an interior point of $f\bigl((x_0-\delta,x_0+\delta)\bigr)$ and therefore an interior point of $f(D)$ and therefore an interior point of $V$.
The case in which $f'(x_0)<0$ is similar.