Find $(f^{-1})''(x)$.

Question

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 $(f^{-1})''$?

Answer 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*}

Answer 2

by the Quotient and Chain rule we get $$(f^{-1})''(x)=-\frac{f''(f^{-1}(x))\cdot (f^{-1}(x))'}{(f'(f^{-1}(x))^2}$$