Quotient rule

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 the low." It is important to remember the 'D' describes the succeeding portion of the original fraction.

Proof

First we have our function:

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

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.

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

Finally, use the least common denominator (which is ${\displaystyle h(x)^{2}}$) to combine the fractions.

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

Implicit differentiation and higher order formulas

The formula can also be derived using implicit differentiation and the product rule. This results in a simpler formula, especially for higher order derivatives. ${\displaystyle f(x)}$ can be implicitly defined as ${\displaystyle f(x)h(x)=g(x).}$ Using the product rule to differentiate this yields ${\displaystyle f'(x)h(x)+f(x)h'(x)=g'(x).}$ Solving for ${\displaystyle f'(x)}$ yields

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

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.