# 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, [itex]f(x)[itex], can be written as

[itex]f(x) = \frac{g(x)}{h(x)}[itex]

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

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

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

[itex]f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{h(a)^2}[itex]
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## Examples

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

[itex]

\frac{d}{dx} \frac{(4x - 2)}{x^2 + 1} [itex]

[itex]

= \frac{(x^2 + 1)(4) - (4x - 2)(2x)}{(x^2 + 1)^2} [itex]

[itex]

= \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} [itex]

[itex]

= \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2} [itex]

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

[itex]

\frac{\cos(x) x^2 - \sin(x)2x}{x^4} [itex]

Another example is:

[itex] f(x) = \frac{2x^2}{x^3}[itex]

whereas [itex]g(x) = 2x^2[itex] and [itex]h(x) = x^3[itex], and [itex]g'(x) = 4x[itex] and [itex]h'(x) = 3x^2[itex].

The derivative of [itex]f(x)[itex] is determined as follows:

[itex]

f'(x) = \frac

{\left[\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)\right]}
{\left(x^3\right)^2}


[itex]

[itex]

= \frac{4x^4 - 6x^4}{x^6} [itex]

[itex]

= \frac{-2x^4}{x^6} [itex]

[itex]

= \frac{-2}{x^2} [itex]

## Proofs

### From Newton's difference quotient

[itex]\mbox{let }f(x) = \frac{g(x)}{h(x)}[itex]
where [itex]h(x) \ne 0[itex] and [itex]g[itex] and [itex]h[itex] are differentiable.
[itex]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}[itex]
[itex]= \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][itex]
[itex]= \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][itex]
[itex]= \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][itex]
[itex]= \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)}[itex]
[itex]= \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))}[itex]
[itex]= \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}[itex]

### From the product rule

[itex]\mbox{let }f(x)=\frac{g(x)}{h(x)}[itex]
[itex]g(x)=f(x)h(x)\mbox{ }[itex]
[itex]g'(x)=f'(x)h(x) + f(x)h'(x)\mbox{ }[itex]

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

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

## 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.

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