Reciprocal rule

Statement

Let $g$ be a real function on an open interval $I$, differentiable at $a \in I$, with $g(a) \ne 0$. Then $1/g$ is differentiable at $a$, and $$\left( \frac{1}{g} \right)'(a) = -\frac{g'(a)}{g(a)^2}.$$

Remarks

The difference quotient is $-\frac{g(a + h) - g(a)}{h} \cdot \frac{1}{g(a + h) g(a)}$. Continuity of $g$ at $a$ gives the limit. For example, $(1/x)' = -1/x^2$ for $x \ne 0$. Mathlib states it as HasDerivAt.inv.

Explanations

L1 human

When $g$ grows, $1/g$ shrinks, so the two derivatives have opposite signs. The factor $1/g^2$ shows that $1/g$ changes fast where $g$ is small.