Let $$ f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \rightarrow \mathbb{R} $$ be defined by $$ f(x) = \frac{x}{\cos x}. $$
We're supposed to show that $f$ has a differentiable inverse $$f^{(-1)}$$ and to calculate $$(f^{-1})'(0)$$
Can we use the mean value theorem to show $f$ has a differentiable inverse? And how would we then calculate the derivative at $x=0$?
This is a straightforward application of the Inverse Function Theorem. The function $f$ is continuously differentiable in $(-\pi/2,\pi/2)$ with $f'(0)=1$. Hence, by the theorem, $f$ is invertible in a neighborhood of $0$, and $f^{-1}$ is continuously differentiable with $(f^{-1})'(0)=1$.