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, <math>f(x)<math>, can be written as
- <math>f(x) = \frac{g(x)}{h(x)}<math>
and <math>h(x) \ne 0<math>, then the rule states that the derivative of <math>g(x)/h(x)<math> is equal to:
- <math>\frac{d}{dx}f(x) = f'(x) = \frac{g'(x)h(x) – g(x)h'(x)}{{h(x)}^2}.<math>
Or more precisely; for all <math>x<math> in some open set containing the number <math>a<math>, with <math>h(a) \ne 0<math>; and, such that <math>g'(a)<math> and <math>h'(a)<math> both exist; then, <math>f'(a)<math> exists as well:
- <math>f'(a)=\frac{g'(a)h(a) – g(a)h'(a)}{h(a)^2}<math>
Table of contents |
Examples
The derivative of <math>\frac{(4x – 2)}{x^2 + 1}<math> is:
- <math>
\frac{d}{dx} \frac{(4x – 2)}{x^2 + 1} <math>
- <math>
= \frac{(x^2 + 1)(4) – (4x – 2)(2x)}{(x^2 + 1)^2} <math>
- <math>
= \frac{(4x^2 + 4) – (8x^2 – 4x)}{(x^2 + 1)^2} <math>
- <math>
= \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2} <math>
The derivative of <math>\frac{\sin(x)}{x^2}<math> (when <math>x \ne 0<math>) is:
- <math>
\frac{\cos(x) x^2 – \sin(x)2x}{x^4} <math>
For more information regarding the derivatives of trigonometric functions, see: derivative.
Another example is:
- <math> f(x) = \frac{2x^2}{x^3}<math>
whereas <math>g(x) = 2x^2<math> and <math>h(x) = x^3<math>, and <math>g'(x) = 4x<math> and <math>h'(x) = 3x^2<math>.
The derivative of <math>f(x)<math> is determined as follows:
- <math>
f'(x) = \frac
{\left[\left(4x \cdot x^3 \right) – \left(2x^2 \cdot 3x^2 \right)\right]}
{\left(x^3\right)^2}
<math>
- <math>
= \frac{4x^4 – 6x^4}{x^6} <math>
- <math>
= \frac{-2x^4}{x^6} <math>
- <math>
= \frac{-2}{x^2} <math>
Proofs
From Newton's difference quotient
- <math>\mbox{let }f(x) = \frac{g(x)}{h(x)}<math>
- where <math>h(x) \ne 0<math> and <math>g<math> and <math>h<math> 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 <math>f'(x)<math> the only term on the left hand side of the equation and to remove <math>f(x)<math> 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.
See also
Categories: Calculus