Formula for the derivative of a ratio of functions
Incalculus , thequotient rule is a method of finding thederivative of afunction that is the ratio of two differentiable functions. Leth ( x ) = f ( x ) g ( x ) {\displaystyle h(x)={\frac {f(x)}{g(x)}}} f andg are differentiable andg ( x ) ≠ 0. {\displaystyle g(x)\neq 0.} h (x )
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 otherderivative rules .
Example 1: Basic example [ edit ] Givenh ( x ) = e x x 2 {\displaystyle h(x)={\frac {e^{x}}{x^{2}}}} f ( x ) = e x , g ( x ) = x 2 {\displaystyle f(x)=e^{x},g(x)=x^{2}} 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 [ edit ] The quotient rule can be used to find the derivative oftan x = sin x cos x {\displaystyle \tan x={\frac {\sin x}{\cos x}}} 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}}}
The reciprocal rule is a special case of the quotient rule in which the numeratorf ( x ) = 1 {\displaystyle f(x)=1} 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 thechain rule yields the same result.
Proof from derivative definition and limit properties [ edit ] Leth ( x ) = f ( x ) g ( x ) . {\displaystyle h(x)={\frac {f(x)}{g(x)}}.} f ( x ) g ( x ) {\displaystyle f(x)g(x)} 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}}} lim k → 0 1 g ( x + k ) g ( x ) = 1 [ g ( x ) ] 2 {\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{[g(x)]^{2}}}} g ( x ) {\displaystyle g(x)} lim k → 0 g ( x + k ) = g ( x ) {\displaystyle \lim _{k\to 0}g(x+k)=g(x)}
Proof using implicit differentiation [ edit ] Leth ( x ) = f ( x ) g ( x ) , {\displaystyle h(x)={\frac {f(x)}{g(x)}},} f ( x ) = g ( x ) h ( x ) . {\displaystyle f(x)=g(x)h(x).}
Theproduct rule then givesf ′ ( x ) = g ′ ( x ) h ( x ) + g ( x ) h ′ ( x ) . {\displaystyle f'(x)=g'(x)h(x)+g(x)h'(x).}
Solving forh ′ ( x ) {\displaystyle h'(x)} h ( x ) {\displaystyle h(x)} 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 [ edit ] Leth ( x ) = f ( x ) g ( x ) = f ( x ) ⋅ 1 g ( x ) . {\displaystyle h(x)={\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}.}
Then the product rule givesh ′ ( x ) = f ′ ( x ) ⋅ 1 g ( x ) + f ( x ) ⋅ d d x [ 1 g ( x ) ] . {\displaystyle 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 thereciprocal rule , or thepower rule along with thechain rule :d d x [ 1 g ( x ) ] = − 1 g ( x ) 2 ⋅ g ′ ( x ) = − g ′ ( x ) g ( x ) 2 . {\displaystyle {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]=-{\frac {1}{g(x)^{2}}}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2}}}.}
Substituting the result into the expression givesh ′ ( x ) = f ′ ( x ) ⋅ 1 g ( x ) + f ( x ) ⋅ [ − g ′ ( x ) g ( x ) 2 ] = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 = g ( x ) g ( x ) ⋅ f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 . {\displaystyle {\begin{aligned}h'(x)&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot \left[{\frac {-g'(x)}{g(x)^{2}}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {g(x)}{g(x)}}\cdot {\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}}
Proof by logarithmic differentiation [ edit ] Leth ( x ) = f ( x ) g ( x ) . {\displaystyle h(x)={\frac {f(x)}{g(x)}}.} absolute value andnatural logarithm of both sides of the equation givesln | h ( x ) | = ln | f ( x ) g ( x ) | {\displaystyle \ln |h(x)|=\ln \left|{\frac {f(x)}{g(x)}}\right|}
Applying properties of the absolute value and logarithms,ln | h ( x ) | = ln | f ( x ) | − ln | g ( x ) | {\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|}
Taking thelogarithmic derivative of both sides,h ′ ( x ) h ( x ) = f ′ ( x ) f ( x ) − g ′ ( x ) g ( x ) {\displaystyle {\frac {h'(x)}{h(x)}}={\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}}
Solving forh ′ ( x ) {\displaystyle h'(x)} f ( x ) g ( x ) {\displaystyle {\tfrac {f(x)}{g(x)}}} h ( x ) {\displaystyle h(x)} h ′ ( x ) = h ( x ) [ f ′ ( x ) f ( x ) − g ′ ( x ) g ( x ) ] = f ( x ) g ( x ) [ f ′ ( x ) f ( x ) − g ′ ( x ) g ( x ) ] = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 . {\displaystyle {\begin{aligned}h'(x)&=h(x)\left[{\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}\right]\\&={\frac {f(x)}{g(x)}}\left[{\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}}
Taking the absolute value of the functions is necessary for thelogarithmic differentiation of functions that may have negative values, as logarithms are onlyreal-valued for positive arguments. This works becaused d x ( ln | u | ) = u ′ u {\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}}
Higher order derivatives [ edit ] Implicit differentiation can be used to compute then th derivative of a quotient (partially in terms of its firstn − 1f = g h {\displaystyle f=gh} f ″ = g ″ h + 2 g ′ h ′ + g h ″ {\displaystyle f''=g''h+2g'h'+gh''} h ″ {\displaystyle h''} h ″ = ( f g ) ″ = f ″ − g ″ h − 2 g ′ h ′ g . {\displaystyle h''=\left({\frac {f}{g}}\right)''={\frac {f''-g''h-2g'h'}{g}}.}
