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] Let h ( x ) = f ( x ) g ( x ) {\displaystyle h(x)={\frac {f(x)}{g(x)}}} , where both f and g are differentiable and g ( x ) ≠ 0. {\displaystyle g(x)\neq 0.} The quotient rule states that the derivative of 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
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Reciprocal rule
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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
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Proof from derivative definition and limit properties
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Let h ( x ) = f ( x ) g ( x ) . {\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 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 {\displaystyle \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 \lim _{k\to 0}g(x+k)=g(x)} .
Proof using implicit differentiation
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Let h ( x ) = f ( x ) g ( x ) , {\displaystyle 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
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Let h ( 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 gives h ′ ( 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 the reciprocal rule , or the power rule along with the chain 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 gives
h ′ ( 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
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Let h ( x ) = f ( x ) g ( x ) . {\displaystyle h(x)={\frac {f(x)}{g(x)}}.} Taking the absolute value and natural logarithm of both sides of the equation gives
ln | 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 the logarithmic 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 for h ′ ( x ) {\displaystyle h'(x)} and substituting back f ( x ) g ( x ) {\displaystyle {\tfrac {f(x)}{g(x)}}} for h ( x ) {\displaystyle h(x)} gives:
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 the logarithmic differentiation of functions that may have negative values, as logarithms are only real-valued for positive arguments. This works because d d x ( ln | u | ) = u ′ u {\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}} , which justifies taking the absolute value of the functions for logarithmic differentiation.
Higher order derivatives
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See also
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References
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