In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let h ( x ) = f ( x ) g ( x ) {\displaystyle \textstyle h(x)={\frac {f(x)}{g(x)}}} , where both f {\displaystyle f} and g {\displaystyle g} are differentiable and g ( x ) ≠ 0 {\displaystyle g(x)\neq 0} . The quotient rule states that the derivative of h ( x ) {\displaystyle h(x)} is
h ′ ( x ) = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) ( g ( x ) ) 2 . {\displaystyle h'(x)={\frac {f'(x)g(x)-f(x)g'(x)}{(g(x))^{2}}}.}
It is provable in many ways by using other derivative rules.
Examples
Example 1: Basic example Given h ( x ) = e x x 2 {\displaystyle \textstyle h(x)={\frac {e^{x}}{x^{2}}}} , let f ( x ) = e x {\displaystyle f(x)=e^{x}} , g ( x ) = x 2 {\displaystyle g(x)=x^{2}} , then using the quotient rule:
d d x ( e x x 2 ) = ( d d x e x ) ( x 2 ) − ( e x ) ( d d x x 2 ) ( x 2 ) 2 = ( e x ) ( x 2 ) − ( e x ) ( 2 x ) x 4 = x 2 e x − 2 x e x x 4 = x e x − 2 e x x 3 = e x ( x − 2 ) x 3 . {\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}}}
Example 2: Derivative of tangent function The quotient rule can be used to find the derivative of tan x = sin x cos x {\textstyle \tan x={\frac {\sin x}{\cos x}}} as follows:
d d x tan x = d d x ( sin x cos x ) = ( d d x sin x ) ( cos x ) − ( sin x ) ( d d x cos x ) cos 2 x = ( cos x ) ( cos x ) − ( sin x ) ( − sin x ) cos 2 x = cos 2 x + sin 2 x cos 2 x = 1 cos 2 x = sec 2 x . {\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}}}
Reciprocal rule
The reciprocal rule is a special case of the quotient rule in which the numerator f ( x ) = 1 {\displaystyle f(x)=1} . Applying the quotient rule gives
h ′ ( x ) = d d x [ 1 g ( x ) ] = 0 ⋅ g ( x ) − 1 ⋅ g ′ ( x ) g ( x ) 2 = − g ′ ( x ) g ( x ) 2 . {\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.
Proofs
Proof from derivative definition and limit properties Let h ( x ) = f ( x ) g ( x ) {\displaystyle \textstyle h(x)={\frac {f(x)}{g(x)}}} . Applying the definition of the derivative and properties of limits gives the following proof, with the term f ( x ) g ( x ) {\displaystyle f(x)g(x)} added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:
h ′ ( x ) = lim k → 0 h ( x + k ) − h ( x ) k = lim k → 0 f ( x + k ) g ( x + k ) − f ( x ) g ( x ) k = lim k → 0 f ( x + k ) g ( x ) − f ( x ) g ( x + k ) k ⋅ g ( x ) g ( x + k ) = lim k → 0 f ( x + k ) g ( x ) − f ( x ) g ( x + k ) k ⋅ lim k → 0 1 g ( x ) g ( x + k ) = lim k → 0 [ f ( x + k ) g ( x ) − f ( x ) g ( x ) + f ( x ) g ( x ) − f ( x ) g ( x + k ) k ] ⋅ 1 [ g ( x ) ] 2 = [ lim k → 0 f ( x + k ) g ( x ) − f ( x ) g ( x ) k − lim k → 0 f ( x ) g ( x + k ) − f ( x ) g ( x ) k ] ⋅ 1 [ g ( x ) ] 2 = [ lim k → 0 f ( x + k ) − f ( x ) k ⋅ g ( x ) − f ( x ) ⋅ lim k → 0 g ( x + k ) − g ( x ) k ] ⋅ 1 [ g ( x ) ] 2 = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 . {\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 lim k → 0 1 g ( x + k ) g ( x ) = 1 [ g ( x ) ] 2 {\textstyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{[g(x)]^{2}}}} is justified by the differentiability of g ( x ) {\displaystyle g(x)} , implying continuity, which can be expressed as lim k → 0 g ( x + k ) = g ( x ) {\displaystyle \textstyle \lim _{k\to 0}g(x+k)=g(x)} .
Proof using implicit differentiation Let h ( x ) = f ( x ) g ( x ) {\displaystyle \textstyle h(x)={\frac {f(x)}{g(x)}}} , so that f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)=g(x)h(x)} . The product rule then gives f ′ ( x ) = g ′ ( x ) h ( x ) + g ( x ) h ′ ( x ) {\displaystyle f'(x)=g'(x)h(x)+g(x)h'(x)} . Solving for h ′ ( x ) {\displaystyle h'(x)} and substituting back for h ( x ) {\displaystyle h(x)} gives:
h ′ ( x ) = f ′ ( x ) − g ′ ( x ) h ( x ) g ( x ) = f ′ ( x ) − g ′ ( x ) ⋅ f ( x ) g ( x ) g ( x ) = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 . {\displaystyle {\begin{aligned}h'(x)&={\frac {f'(x)-g'(x)h(x)}{g(x)}}\\&={\frac {f'(x)-g'(x)\cdot {\frac {f(x)}{g(x)}}}{g(x)}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}.\end{aligned}}}
Proof using the reciprocal rule or chain rule Let h ( x ) = f ( x ) g ( x ) = f ( x ) ⋅ 1 g ( x ) {\displaystyle \textstyle h(x)={\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}} . Then the product rule gives h ′ ( x ) = f ′ ( x ) ⋅ 1 g ( x ) + f ( x ) ⋅ d d x [ 1 g ( x ) ] {\displaystyle \textstyle h'(x)=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]} . To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule:
d d x [ 1 g