# 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

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

## Examples

1. A basic example:
{\displaystyle {\begin{aligned}{\frac {d}{dx}}{\frac {e^{x}}{x^{2}}}&={\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 {e^{x}(x-2)}{x^{3}}}.\end{aligned}}}
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{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}{\frac {\sin x}{\cos x}}\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}}

## 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 {\begin{aligned}f'(x)&=\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {g(x+k)}{h(x+k)}}-{\frac {g(x)}{h(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k\cdot h(x)h(x+k)}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{h(x)h(x+k)}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)+g(x)h(x)-g(x)h(x+k)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)}{k}}-\lim _{k\to 0}{\frac {g(x)h(x+k)-g(x)h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(h(x)\lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}-g(x)\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}

### Proof using implicit differentiation

Let ${\displaystyle f(x)={\frac {g(x)}{h(x)}},}$ 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 {\begin{aligned}f'(x)&={\frac {g'(x)-f(x)h'(x)}{h(x)}}\\&={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}

### Proof using the chain rule

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

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

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 {\begin{aligned}f'(x)&={\frac {g'(x)}{h(x)}}-{\frac {g(x)h'(x)}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}

## 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 yields ${\displaystyle f''h+2f'h'+fh''=g'',}$ and then solving for ${\displaystyle f''}$ yields

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

## 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.