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.

Encyclopedia

Dictionary

Quotes

Quotient rule

Examples

Proofs

From Newton's Difference Quotient

From the Product Rule

Mnemonic

See also

The Online Encyclopedia and Dictionary

Encyclopedia

Dictionary

Quotes

Quotient rule

Examples

Proofs

From Newton's Difference Quotient

From the Product Rule

Mnemonic

See also