In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}}}$, where both f and g are differentiable and ${\displaystyle g(x)\neq 0.}$ The quotient rule states that the derivative of h(x) is

It is provable in many ways by using other derivative rules.

Given ${\displaystyle h(x)={\frac {e^{x}}{x^{2}}}}$, let ${\displaystyle f(x)=e^{x},g(x)=x^{2}}$, then using the quotient rule:$${\displaystyle {\begin{aligned}{\frac {d}{dx}}\left({\frac {e^{x}}{x^{2}}}\right)&={\frac {\left({\frac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\frac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\frac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\frac {x^{2}e^{x}-2xe^{x}}{x^{4}}}\\&={\frac {xe^{x}-2e^{x}}{x^{3}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}.\end{aligned}}}$$

The quotient rule can be used to find the derivative of ${\displaystyle \tan x={\frac {\sin x}{\cos x}}}$ as follows: $${\displaystyle {\begin{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}\left({\frac {\sin x}{\cos x}}\right)\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}}$$

The reciprocal rule is a special case of the quotient rule in which the numerator ${\displaystyle f(x)=1}$. Applying the quotient rule gives$${\displaystyle h'(x)={\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]={\frac {0\cdot g(x)-1\cdot g'(x)}{g(x)^{2}}}={\frac {-g'(x)}{g(x)^{2}}}.}$$

Utilizing the chain rule yields the same result.

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}}.}$ Applying the definition of the derivative and properties of limits gives the following proof, with the term ${\displaystyle f(x)g(x)}$ added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:$${\displaystyle {\begin{aligned}h'(x)&=\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {f(x+k)}{g(x+k)}}-{\frac {f(x)}{g(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x+k)}{k\cdot g(x)g(x+k)}}\\&=\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{g(x)g(x+k)}}\\&=\lim _{k\to 0}\left[{\frac {f(x+k)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+k)}{k}}\right]\cdot {\frac {1}{[g(x)]^{2}}}\\&=\left[\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x)}{k}}-\lim _{k\to 0}{\frac {f(x)g(x+k)-f(x)g(x)}{k}}\right]\cdot {\frac {1}{[g(x)]^{2}}}\\&=\left[\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\cdot g(x)-f(x)\cdot \lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}\right]\cdot {\frac {1}{[g(x)]^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}.\end{aligned}}}$$The limit evaluation ${\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{[g(x)]^{2}}}}$ is justified by the differentiability of ${\displaystyle g(x)}$, implying continuity, which can be expressed as ${\displaystyle \lim _{k\to 0}g(x+k)=g(x)}$.

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}},}$ so that ${\displaystyle f(x)=g(x)h(x).}$