Quotient rule

Topics in Calculus
Differential calculus
	Derivative; Change of variables; Implicit differentiation; Taylor's theorem; Related rates; Identities; Rules:; Power rule, Product rule, Quotient rule, Chain rule
Integral calculus
	Integral; Lists of integrals; Improper integrals; Integration by:; parts, disks, cylindrical; shells, substitution,; trigonometric substitution,; partial fractions, changing order
Vector calculus
	Gradient; Divergence; Curl; Laplacian; Gradient theorem; Green's theorem; Stokes' theorem; Divergence theorem
Multivariable calculus
	Matrix calculus; Partial derivative; Multiple integral; Line integral; Surface integral; Volume integral; Jacobian

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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 $h(x)\not=0$ , then the rule states that the derivative of $g (x) / h (x)$ is

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

More precisely, if all x in some open set containing the number a satisfy $h(x)\not=0$ , and $g'(a)$ and $h'(a)$ both exist, then $f'(a)$ exists as well and

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

[edit] Examples

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

$\begin{align}\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(x^2 + 1)(4) - (4x - 2)(2x)}{(x^2 + 1)^2}\\&= \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} &= \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align}$

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:

$\frac{\cos(x) x^2 - \sin(x)2x}{x^4}$

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:

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

This can be checked by using laws of exponents and the power rule:

$f(x) = \frac{2x^2}{x^3} = \frac{2}{x} = 2x^{-1}$

$f'(x) = -2x^{-2} = -\frac{2}{x^2}$

[edit] Limitations

The quotient rule is not useful at points where either the numerator or denominator are not differentiable; it's possible that the quotient may be differentiable at such points. For example, consider the function:

$f(x) = \frac{|x|+1}{|x|+1},$

where |x| denotes the absolute value of x. This is of course simply the function f(x) = 1, so it is differentiable everywhere and in particular f'(0) = 0. If we try to use the quotient rule to compute f'(0), however, an undefined value will result, since |x| is nondifferentiable at x = 0.

[edit] Proofs

[edit] From Newton's difference quotient

Suppose $f (x) = g (x) / h (x)$ where $h(x) \neq 0$ and $g$ and $h$ are differentiable.

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

We pull out the $1 / Δ x$ and combine the fractions in the numerator:

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

Adding and subtracting $g (x) h (x)$ in the numerator:

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

We factor this and multiply the $1 / Δ x$ through the numerator:

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

Now we move the limit through:

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

By the definition of the difference quotient, the limits in the numerator are derivatives, so we have:

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

[edit] Using the Chain Rule

Consider the identity

$\frac{u}{v}\; =\; \frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]$

Then

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{d}{dx}\frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]$

Leading to

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{1}{4}\left[ 2\left( u+\frac{1}{v} \right)\left( \frac{du}{dx}-\frac{dv}{v^{2}dx} \right)-\; 2\left( u-\frac{1}{v} \right)\left( \frac{du}{dx}+\frac{dv}{v^{2}dx} \right) \right]$

Multiplying out leads to

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{1}{4}\left[ \frac{4}{v}\frac{du}{dx}-\frac{4u}{v^{2}}\frac{dv}{dx} \right]$

Finally, taking a common denominator leaves us with the expected result

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{\left[ v\frac{du}{dx}-u\frac{dv}{dx} \right]}{v^{2}}$

[edit] By total differentials

An even more elegant proof is a consequence of the 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.

[edit] See also

Quotient rule

From Wikipedia, the free encyclopedia

Contents

[edit] Examples

[edit] Limitations

[edit] Proofs

[edit] From Newton's difference quotient

[edit] Using the Chain Rule

[edit] By total differentials

[edit] See also

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