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Quotient rule

## Summary

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

${\displaystyle h'(x)={\frac {f'(x)g(x)-f(x)g'(x)}{(g(x))^{2}}}.}$

It is provable in many ways by using other derivative rules.

## Examples

### Example 1: Basic example

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

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

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

${\displaystyle h'(x)={\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]={\frac {0\cdot g(x)-1\cdot g'(x)}{g(x)^{2}}}={\frac {-g'(x)}{g(x)^{2}}}.}$

Utilizing the chain rule yields the same result.

## Proofs

### Proof from derivative definition and limit properties

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

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

### Proof using implicit differentiation

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

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

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

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

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

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

${\displaystyle {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]=-{\frac {1}{g(x)^{2}}}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2}}}.}$

Substituting the result into the expression gives

{\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

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

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

Applying properties of the absolute value and logarithms,

${\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|}$

Taking the logarithmic derivative of both sides,

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

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

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

## Higher order derivatives

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

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