Quotient rule

In calculus, the quotient rule is a method of finding the derivative of a function which is the quotient of two other functions for which derivatives exist.

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

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

and $h(x) \ne 0$ , then the rule states that the derivative of $g (x) / h (x)$ is equal to:

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

Or more precisely; for all $x$ in some open set containing the number $a$ , with $h(a) \ne 0$ ; and, such that $g'(a)$ and $h'(a)$ both exist; then, $f'(a)$ exists as well:

$f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{h(a)^2}$

Contents

1 Examples

2 Proofs

2.1 From Newton's difference quotient
2.2 From the product rule

3 Mnemonic

4 See also

Examples

The derivative of $\frac{(4x - 2)}{x^2 + 1}$ is:

$\frac{d}{dx} \frac{(4x - 2)}{x^2 + 1}$

$= \frac{(x^2 + 1)(4) - (4x - 2)(2x)}{(x^2 + 1)^2}$

$= \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2}$

$= \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}$

The derivative of $\frac{\sin(x)}{x^2}$ (when $x \ne 0$ ) is:

$\frac{\cos(x) x^2 - \sin(x)2x}{x^4}$

For more information regarding the derivatives of trigonometric functions, see: derivative.

Another example is:

$f(x) = \frac{2x^2}{x^3}$

whereas $g (x) = 2 x 2$ and $h (x) = x 3$ , and $g'(x) = 4 x$ and $h'(x) = 3 x 2$ .

The derivative of $f (x)$ is determined as follows:

$f'(x) = \frac {\left[\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)\right]} {\left(x^3\right)^2}$

$= \frac{4x^4 - 6x^4}{x^6}$

$= \frac{-2x^4}{x^6}$

$= \frac{-2}{x^2}$

Proofs

From Newton's difference quotient

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

where $h(x) \ne 0$ and

g

and

h

are differentiable.

$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{g(x + \Delta x)}{h(x + \Delta x)} - \frac{g(x)}{h(x)}}{\Delta x}$

$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \frac{g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)} \right]$

$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \frac{[g(x+\Delta x)h(x)-g(x)h(x)]-[g(x)h(x+\Delta x)-g(x)h(x)]}{h(x)h(x+\Delta x)} \right]$

$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \frac{h(x)[g(x+\Delta x)-g(x)]-g(x)[h(x+\Delta x)-h(x)]}{h(x)h(x+\Delta x)} \right]$

$= \lim_{\Delta x \to 0} \frac{\frac{g(x+\Delta x)-g(x)}{\Delta x}h(x)-g(x)\frac{h(x+\Delta x)-h(x)}{\Delta x}}{h(x)h(x+\Delta x)}$

$= \frac{\lim_{\Delta x \to 0} \left(\frac{g(x+\Delta x)-g(x)}{\Delta x}\right)h(x) - g(x) \lim_{\Delta x \to 0} \left(\frac{h(x+\Delta x)-h(x)}{\Delta x}\right)}{h(x)h(\lim_{\Delta x \to 0} (x+\Delta x))}$

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

From the product rule

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

g (x) = f (x) h (x)

g'(x) = f'(x) h (x) + f (x) h'(x)

The rest is simple algebra to make $f'(x)$ the only term on the left hand side of the equation and to remove $f (x)$ from the right side of the equation.

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

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

Mnemonic

It is often memorized as a rhyme type song. "Lo-dee-hi minus hi-dee-lo all over lo-lo"; Lo being the denominator, Hi being the numerator and D being the derivative.