Quotient rule

From Wikipedia, the free encyclopedia

In calculus, the quotient rule is a method of finding the derivative of a function that 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 $h (x)$ ≠ $0$ , 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 $h (a)$ ≠ $0$ ; and, such that $g'(a)$ and $h'(a)$ both exist; then, $f'(a)$ exists as well:

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

1 Examples
2 Proofs
3 Mnemonic
4 See also

The derivative of $(4 x - 2) / (x 2 + 1)$ is:

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

In the example above, the choices

g (x) = 4 x - 2

h (x) = x 2 + 1

were made. Analogously, the derivative of $sin(x) / x 2$ (when $x$ ≠ 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:

$f'(x)\,$	$=\frac {\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)} {\left(x^3\right)^2}$
	$=\frac{4x^4 - 6x^4}{x^6}$
	$=\frac{-2x^4}{x^6}$
	$=-\frac{2}{x^2}$

Suppose

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

where

h (x)

≠ 0 and

g

and

h

are differentiable.

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

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

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

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

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

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

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

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

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

$g'(x)=f'(x)h(x) + f(x)h'(x)\mbox{ } \,$

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.

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

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

Alternatively, we can just apply the product rule directly, without having to use substitution:

$f(x) = \frac{g(x)}{h(x)} = g(x) [h(x)]^{-1}$

Followed by using the chain rule to differentiate $h (x) - 1$ :

$f'(x) = g'(x) [h(x)]^{-1} + g(x) (-1) [h(x)]^{-2} h'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}$

An even more elegant proof is a consequence of the old law about total differentials, which states that the total differential,

$dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy + \frac{\partial F}{\partial z} dz + ...$

of any function in any set of quantities is decomposable in this way, no matter what the independent variables in a function are (i.e., no matter which variables are taken so that they may not be expressed as functions of other variables). This means that, if N and D are both functions of an independent variable x, and $F = N (x) / D (x)$ , then it must be true both that

(*) $dF = \frac{\partial F}{\partial x}dx$

and that

$dF = \frac{\partial F}{\partial N}dN + \frac{\partial F}{\partial D}dD$ .

But we know that $d N = N'(x) d x$ and $d D = D'(x) d x$ .

Substituting and setting these two total differentials equal to one another (since they represent limits which we can manipulate), we obtain the equation

$\frac{\partial F}{\partial x} dx = \frac{\partial F}{\partial N}N'(x) dx + \frac{\partial F}{\partial D}D'(x) dx$

which requires that

(#) $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial N}N'(x) + \frac{\partial F}{\partial D}D'(x)$ .

We compute the partials on the right:

$\frac{\partial F}{\partial N} = \frac{\partial (N/D)}{\partial N} = \frac{1}{D}$ ;

$\frac{\partial F}{\partial D} = \frac{\partial (N/D)}{\partial D} = -\frac{N}{D^2}$ .

If we substitute them into (#),

$\frac{\partial F}{\partial x} = \frac{N'(x)}{D(x)} - \frac{N(x) D'(x)}{D(x)^2}$

$\frac{\partial F}{\partial x} = \frac{D(x)N'(x)}{D(x)^2} - \frac{N(x) D'(x)}{D(x)^2}$

which gives us the quotient rule, since, by (*),

$\frac{dF}{dx} = \frac{\partial F}{\partial x}$ .

This proof, of course, is just another, more systematic (even if outmoded) way of proving the theorem in terms of limits, and is therefore equivalent to the first proof above - and even reduces to it, if you make the right substitutions in the right places. Students of multivariable calculus will recognize it as one of the chain rules for functions of multiple variables.

It is often memorized as a rhyme type song. "Lo-dee-hi, hi-dee-lo, draw the line and square below"; Lo being the denominator, Hi being the numerator and "dee" being the derivative. Another variation to this mnemonic is given when the quotient is written with the numerator as Hi the denominator as Ho: "Ho-dee-Hi minus Hi-dee-Ho all over Ho-Ho." A third variation is "Low-dee-high minus high-dee-low, all over the square of what's below".

Retrieved from "http://en.wikipedia.org../../../q/u/o/Quotient_rule.html"

Category: Differential calculus

Searching reference...

Reference

Sources

Quotient rule

From Wikipedia, the free encyclopedia

Contents

Top Matching Results

Sponsored Links
This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.

Search Results

Searching reference...

Reference

Sources

Quotient rule

From Wikipedia, the free encyclopedia

Contents

Top Matching Results

Sponsored Links This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.

Search Results

Sponsored Links
This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.