In calculus, the reciprocal rule gives the derivative of the reciprocal of a function f in terms of the derivative of f. The reciprocal rule can be used to show that the power rule holds for negative exponents if it has already been established for positive exponents. Also, one can readily deduce the quotient rule from the reciprocal rule and the product rule.

The reciprocal rule states that if f is differentiable at a point x and f(x) ≠ 0 then g(x) = 1/f(x) is also differentiable at x and

$${\displaystyle g'(x)=-{\frac {f'(x)}{f(x)^{2}}}.}$$

This proof relies on the premise that ${\displaystyle f}$ is differentiable at ${\displaystyle x,}$ and on the theorem that ${\displaystyle f}$ is then also necessarily continuous there. Applying the definition of the derivative of ${\displaystyle g}$ at ${\displaystyle x}$ with ${\displaystyle f(x)\neq 0}$ gives $${\displaystyle {\begin{aligned}g'(x)={\frac {d}{dx}}\left({\frac {1}{f(x)}}\right)&=\lim _{h\to 0}\left({\frac {{\frac {1}{f(x+h)}}-{\frac {1}{f(x)}}}{h}}\right)\\&=\lim _{h\to 0}\left({\frac {f(x)-f(x+h)}{h\cdot f(x)\cdot f(x+h)}}\right)\\&=\lim _{h\to 0}\left(-({\frac {(f(x+h)-f(x)}{h}})\cdot ({\frac {1}{f(x)\cdot f(x+h)}})\right)\end{aligned}}}$$ The limit of this product exists and is equal to the product of the existing limits of its factors: $${\displaystyle \left(-\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\right)\cdot \left(\lim _{h\to 0}{\frac {1}{f(x)\cdot f(x+h)}}\right)}$$ Because of the differentiability of ${\displaystyle f}$ at ${\displaystyle x}$ the first limit equals ${\displaystyle -f'(x),}$ and because of ${\displaystyle f(x)\neq 0}$ and the continuity of ${\displaystyle f}$ at ${\displaystyle x}$ the second limit equals ${\displaystyle 1/f(x)^{2},}$ thus yielding $${\displaystyle g'(x)=-f'(x)\cdot {\frac {1}{f(x)^{2}}}=-{\frac {f'(x)}{f(x)^{2}}}}$$

It may be argued that since

$${\displaystyle f(x)\cdot {\frac {1}{f(x)}}=1,}$$

an application of the product rule says that

$${\displaystyle f'(x)\left({\frac {1}{f}}\right)(x)+f(x)\left({\frac {1}{f}}\right)'(x)=0,}$$