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

${\displaystyle f'(x)={\frac {[h(x)]^}}.}$

## Examples

1. A basic example:
${\displaystyle {\begin{\frac }{\frac }}}&={\frac {\left({\frac }e^\right)(x^)-(e^)\left({\frac }x^\right)}{(x^)^}}\\&={\frac {(e^)(x^)-(e^)(2x)}}}\\&={\frac (x-2)}}}.\end}}$
2. The quotient rule can be used to find the derivative of ${\displaystyle f(x)=\tan x={\tfrac {\sin x}{\cos x}}}$ as follows.
${\displaystyle {\begin{\frac }\tan x&={\frac }{\frac {\sin x}{\cos x}}\\&={\frac {\left({\frac }\sin x\right)(\cos x)-(\sin x)\left({\frac }\cos x\right)}{\cos ^x}}\\&={\frac {\cos ^x+\sin ^x}{\cos ^x}}\\&={\frac {\cos ^x}}=\sec ^x.\end}}$

## Proofs

### Proof from derivative definition and limit properties

Let ${\displaystyle f(x)=g(x)/h(x).}$ Applying the definition of the derivative and properties of limits gives the following proof.

${\displaystyle {\beginf'(x)&=\lim _{\frac }\\&=\lim _{\frac {{\frac }-{\frac }}}\\&=\lim _{\frac }\\&=\lim _{\frac }\cdot \lim _{\frac }\\&=\left(\lim _{\frac }\right)\cdot {\frac }}\\&=\left(\lim _{\frac }-\lim _{\frac }\right)\cdot {\frac }}\\&=\left(h(x)\lim _{\frac }-g(x)\lim _{\frac }\right)\cdot {\frac }}\\&={\frac }}.\end}}$

### Proof using implicit differentiation

Let ${\displaystyle f(x)={\frac },}$ so ${\displaystyle g(x)=f(x)h(x).}$ The product rule then gives ${\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x).}$ Solving for ${\displaystyle f'(x)}$ and substituting back for ${\displaystyle f(x)}$ gives:

${\displaystyle {\beginf'(x)&={\frac }\\&={\frac }\cdot h'(x)}}\\&={\frac }}.\end}}$

### Proof using the chain rule

Let ${\displaystyle f(x)={\frac }=g(x)h(x)^{-1}.}$ Then the product rule gives

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

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

${\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).}$

Finally, rewrite as fractions and combine terms to get

${\displaystyle {\beginf'(x)&={\frac }-{\frac }}\\&={\frac }}.\end}}$

## Higher order formulas

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 ${\displaystyle fh=g}$ twice (resulting in ${\displaystyle f''h+2f'h'+fh''=g''}$) and then solving for ${\displaystyle f''}$ yields

${\displaystyle f''=\left({\frac }\right)''={\frac }.}$

## References

1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5.
2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 0-547-16702-4.
3. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 0-321-58876-2.