# Quotient rule

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In calculus, the quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives exist.[1][2][3]

If the function one wishes to differentiate, ${\displaystyle f(x)}$, can be written as

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

and ${\displaystyle h(x)\not =0}$, then the rule states that the derivative of ${\displaystyle g(x)/h(x)}$ is

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

Many people remember the Quotient Rule by the rhyme "Low D-high, High D-low, cross the line and square below." It is important to remember the 'D' describes the succeeding portion of the original fraction.

## Proof using implicit differentiation

Let ${\displaystyle f(x)={\frac {g(x)}{h(x)}}}$
Then ${\displaystyle g(x)=f(x)h(x){\mbox{ }}\,}$
Using product rule, ${\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x){\mbox{ }}\,}$
${\displaystyle f'(x)={\frac {g'(x)-f(x)h'(x)}{h(x)}}={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}}$
${\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{\left(h(x)\right)^{2}}}}$

## Proof using chain rule

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

We rewrite the fraction using a negative exponent.

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

Take the derivative of both sides, and apply the product rule to the right side.

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

To evaluate the derivative in the second term, apply the chain rule, where the outer function is ${\displaystyle x^{-1}}$, and the inner function is ${\displaystyle h(x)}$.

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

Rewrite things in fraction form.

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

## Higher order formulas

It is much easier to derive higher order quotient rules using implicit differentiation. For example, two implicit differentiations of ${\displaystyle fh=g}$ yields ${\displaystyle f''h+2f'h'+fh''=g''}$ and solving for ${\displaystyle f''}$ yields

${\displaystyle f''={\frac {g''-2f'h'-fh''}{h}}.}$

## 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.