#math/calculus If $g$ is differentiable and $g(x)\neq 0$, then $ \left(\frac{1}{g}\right)'(x)=-\frac{g'(x)}{[g(x)]^{2}}. $ Indeed, $g(x)\cdot[1/g(x)]=1$. Differentiating with the product rule gives $ g'(x)\frac{1}{g(x)}+g(x)\left(\frac{1}{g}\right)'(x)=0, $ and solving for the unknown derivative yields the rule.