Reciprocal Rule for One-Dimensional Derivatives

Let $\mathbb{R}$ be the real numbers, an ordered field; write $z^{-1}$ for the multiplicative inverse of $z\in\mathbb{R}$ with $z\ne 0$, and $w^{2}$ for $w\cdot w$. Let $I\subseteq\mathbb{R}$ be an interval and let $x_{0}\in I$ be an interior point of $I$; differentiability at $x_{0}$ is that of Derivative at an Interior Point.

Then the following hold.

1. (Reciprocal rule) Let $g:I\to\mathbb{R}$ satisfy $g(z)\ne 0$ for every $z\in I$, and let $1/g:I\to\mathbb{R}$ be the function whose value at $z\in I$ is $g(z)^{-1}$. If $g$ is differentiable at $x_{0}$, then $1/g$ is differentiable at $x_{0}$, and

$$(1/g)'(x_{0})=-\,g'(x_{0})\,\bigl(g(x_{0})^{-1}\bigr)^{2}.$$

2. (The reciprocal map) Suppose $0\notin I$, and let $r:I\to\mathbb{R}$ be the function whose value at $z\in I$ is $z^{-1}$. Then $r$ is differentiable at $x_{0}$, and

$$r'(x_{0})=-\bigl(x_{0}^{-1}\bigr)^{2}.$$