Derivative of an inverse function

Statement

Let $I$ be an open interval and let $f \colon I \to \mathbb{R}$ be continuous and strictly monotone. Assume that $f$ is differentiable at $a \in I$ with $f'(a) \ne 0$, and let $b = f(a)$. Then $f^{-1}$ is differentiable at $b$, and $$(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}.$$

Remarks

The inverse is continuous, so $f^{-1}(b + k) \to a$ as $k \to 0$. The difference quotient of $f^{-1}$ at $b$ is the reciprocal of a difference quotient of $f$ at $a$. If $f'(a) = 0$, then $f^{-1}$ is not differentiable at $b$. An example is $f(x) = x^3$ at $a = 0$. For example, it gives the derivative of the logarithm from the derivative of the exponential function.

Explanations

L1 human

The graph of $f^{-1}$ is the graph of $f$ reflected in the line $y = x$. The reflection turns a slope $m$ into the slope $1/m$.