Quotient rule

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 $h(x)={\frac {f(x)}{g(x)}}$ , where both $f$ and $g$ are differentiable and $g(x)\neq 0.$ The quotient rule states that the derivative of $h (x)$ is

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 $h(x)={\frac {e^{x}}{x^{2}}}$ , let $f(x)=e^{x},g(x)=x^{2}$ , then using the quotient rule:

{\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 $\tan x={\frac {\sin x}{\cos x}}$ as follows:

{\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 $f(x)=1$ . Applying the quotient rule gives

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}}}.

Note that utilizing the chain rule yields the same result.

Proofs

Proof from derivative definition and limit properties

Let $h(x)={\frac {f(x)}{g(x)}}.$ Applying the definition of the derivative and properties of limits gives the following proof, with the term $f(x)g(x)$ added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:

{\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 $\lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{g(x)^{2}}}$ is justified by the differentiability of $g(x)$ , implying continuity, which can be expressed as $\lim _{k\to 0}g(x+k)=g(x)$ .

Proof using implicit differentiation

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

The product rule then gives $f'(x)=g'(x)h(x)+g(x)h'(x).$

Solving for $h'(x)$ and substituting back for $h(x)$ gives:

{\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 $h(x)={\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}.$

Then the product rule gives $h'(x)=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right].$

To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule:

{\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

{\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

Let $h(x)={\frac {f(x)}{g(x)}}.$ Taking the absolute value and natural logarithm of both sides of the equation gives

\ln |h(x)|=\ln \left|{\frac {f(x)}{g(x)}}\right|

Applying properties of the absolute value and logarithms,

\ln |h(x)|=\ln |f(x)|-\ln |g(x)|

Taking the logarithmic derivative of both sides,

{\frac {h'(x)}{h(x)}}={\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}

Solving for $h'(x)$ and substituting back ${\tfrac {f(x)}{g(x)}}$ for $h(x)$ gives:

{\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}}

Note: 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 ${\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}$ , which justifies taking the absolute value of the functions for logarithmic differentiation.

Higher order derivatives

Implicit differentiation can be used to compute the $n$ th derivative of a quotient (partially in terms of its first $n - 1$ derivatives). For example, differentiating $f=gh$ twice (resulting in $f''=g''h+2g'h'+gh''$ ) and then solving for $h''$ yields

h''=\left({\frac {f}{g}}\right)''={\frac {f''-g''h-2g'h'}{g}}.

References

Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.
Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.
Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 978-0-321-58876-0.

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[1] Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.

[2] Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 978-0-547-16702-2.

[3] Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 978-0-321-58876-0.

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