In calculus , the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions.[ 1] [ 2] [ 3]
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{\displaystyle h(x)={\frac {f(x)}{g(x)}}}
f and g are differentiable and
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{\displaystyle g(x)\neq 0.}
h (x )
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{\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 .
The reciprocal rule is a special case of the quotient rule in which the numerator
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{\displaystyle f(x)=1}
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{\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.
Proof from derivative definition and limit properties
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Let
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{\displaystyle h(x)={\frac {f(x)}{g(x)}}.}
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{\displaystyle f(x)g(x)}
h
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{\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
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{\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{[g(x)]^{2}}}}
g
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{\displaystyle g(x)}
lim
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{\displaystyle \lim _{k\to 0}g(x+k)=g(x)}
Proof using implicit differentiation
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Let
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{\displaystyle h(x)={\frac {f(x)}{g(x)}},}
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{\displaystyle f(x)=g(x)h(x).}
The product rule then gives
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{\displaystyle f'(x)=g'(x)h(x)+g(x)h'(x).}
Solving for
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{\displaystyle h'(x)}
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{\displaystyle h(x)}
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{\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
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Let
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{\displaystyle h(x)={\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}.}
Then the product rule gives
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{\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 the reciprocal rule , or the power rule along with the chain rule :
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{\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 gives
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{\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
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Let
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{\displaystyle h(x)={\frac {f(x)}{g(x)}}.}
absolute value and natural logarithm of both sides of the equation gives
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{\displaystyle \ln |h(x)|=\ln \left|{\frac {f(x)}{g(x)}}\right|}
Applying properties of the absolute value and logarithms,
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{\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|}
Taking the logarithmic derivative of both sides,
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{\displaystyle {\frac {h'(x)}{h(x)}}={\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}}
Solving for
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{\displaystyle h'(x)}
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{\displaystyle {\tfrac {f(x)}{g(x)}}}
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{\displaystyle h(x)}
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{\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 the logarithmic differentiation of functions that may have negative values, as logarithms are only real-valued for positive arguments. This works because
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{\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}}
Higher order derivatives
edit
![{\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}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4e609f5e7b44f69edc0c1030bc9ae9d0cb2a104)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a67ddfcb5e8f45f7cc7224aea48f8624bed81c14)
![{\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{[g(x)]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54648c1bbdf7de8fb2914518e64c9d3664cec574)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/178b7215193de54049ec971bbf3d2ee99ce7fe3c)
![{\displaystyle h'(x)=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4eb1ec4c4cd3ce8891ece4a7f4db407fd158470c)
![{\displaystyle {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]=-{\frac {1}{g(x)^{2}}}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3578de83a411d6ecc2753ffb336bbefa818b2508)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98814d0c465bffd44efe44e56bb1a84dd7fb1529)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a30778dbc5e1cd7b1f21fec351752fc348cb88af)
