Quotient rule - Definition and Overview

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

<math>f(x) = \frac{g(x)}{h(x)}<math>

and h(x) ≠ 0; then, the rule states that the derivative of g(x) / h(x) is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator:

<math>f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{{h(x)}^2}.<math>

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

<math>f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{h(a)^2}<math>

Examples

The derivative of (4x - 2) / (x² + 1) = [(x² + 1)(4) - (4x - 2)(2x)] / (x² + 1)² = [(4x² + 4) - (8x² - 4x)] / (x² + 1)² = [-4x² + 4x + 4] / (x² + 1)²

The derivative of [sin(x)] / x² (when x ≠ 0) is ([cos(x)]x² - [sin(x)](2x)) / x⁴.

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

Another example is:

<math> f(x) = {2x^2 \over x^3} <math>

whereas g(x) = 2x² and h(x) = x³, and g′(x) = 4x and h′(x) = 3x².

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

<math> f'(x) = {\left[\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)\right] \over \left(x^3\right)^2} <math>

<math> f'(x) = {4x^4 - 6x^4 \over x^6} <math>

<math> f'(x) = {-2x^4 \over x^6} <math>

<math> f'(x) = {-2 \over x^2} <math>

Proofs

From Newton's Difference Quotient

<math>\mbox{let }f(x) = \frac{g(x)}{h(x)}<math>

where h(x) ≠ 0 and g and h are differentiable.

<math>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}<math>

<math>= \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]<math>

<math>= \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]<math>

<math>= \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]<math>

<math>= \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)}<math>

<math>= \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))}<math>

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

From the Product Rule

<math>\mbox{let }f(x)=\frac{g(x)}{h(x)}<math>

<math>g(x)=f(x)h(x)\mbox{ }<math>

<math>g'(x)=f'(x)h(x) + f(x)h'(x)\mbox{ }<math>

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.

<math>f'(x)=\frac{g'(x) - f(x)h'(x)}{h(x)} = \frac{g'(x) - \frac{g(x)}{h(x)}\cdot h'(x)}{h(x)}<math>

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

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.