Quotient rule

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Calculus
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Differential Definitions Derivative (generalizations) Differential infinitesimal of a function total Concepts Differentiation notation Second derivative Third derivative Change of variables Implicit differentiation Related rates Taylor's theorem Rules and identities Sum Product Chain Power Quotient General Leibniz Faà di Bruno's formula
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Specialized Fractional Malliavin Stochastic Variations

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.

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

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

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

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

Then $g(x)=f(x)h(x){\mbox{ }}\,$$g(x)=f(x)h(x){\mbox{ }}\,$

Using product rule, $g'(x)=f'(x)h(x)+f(x)h'(x){\mbox{ }}\,$$g'(x)=f'(x)h(x)+f(x)h'(x){\mbox{ }}\,$

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

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

Proof using chain rule

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

We rewrite the fraction using a negative exponent.

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

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

$f'(x)=g'(x)\cdot h(x)^{-1}+g(x)\cdot (h(x)^{-1})'$$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 $x^{-1}$$x^{-1}$, and the inner function is $h(x)$$h(x)$.

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

Rewrite things in fraction form.

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

$f'(x)={\frac {g'(x)\cdot h(x)-h'(x)\cdot g(x)}{[h(x)]^{2}}}$$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 $fh=g$$fh=g$ yields $f''h+2f'h'+fh''=g''$$f''h+2f'h'+fh''=g''$ and solving for $f''$$f''$ yields

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