Find $(f^{-1})''(x)$. If $f$ is strictly monotonous and differentiable for a given interval, we have
$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$
How much is $(f^{-1})''$?

\begin{align*}
  y &= f(x) \\
  x &= f^{-1} (y) \\
  f'(x) &= \frac{dy}{dx} \\
   &= \frac{1}{\frac{dx}{dy}} \\
  f''(x) &= \frac{d^{2}y}{dx^{2}} \\
  &= \frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) \times \frac{dy}{dx} \\
  &= -\frac{\frac{d^{2}x}{dy^{2}}}{\left( \frac{dx}{dy} \right)^{3}} \\
   \frac{d^{2}x}{dy^{2}} &=
  -\frac{\frac{d^{2}y}{dx^{2}}}
        {\left( \frac{dy}{dx} \right)^{3}} \\
  [f^{-1}(y)]'' &= -\frac{f''(x)}{[f'(x)]^{3}} \\
  &= -\frac{f''[f^{-1}(y)]}{\{f'[f^{-1}(y)]\}^{3}}
\end{align*}
