Quotient rule
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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, <math>f(x)</math>, can be written as
- <math>f(x) = \frac{g(x)}{h(x)}</math>
and <math>h(x)</math> ≠ <math>0</math>, then the rule states that the derivative of <math>g(x)/h(x)</math> is equal to:
- <math>f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.</math>
Or, more precisely, if all x in some open set containing the number a satisfy <math>h(x)</math> ≠ <math>0</math>; and <math>g'(a)</math> and <math>h'(a)</math> both exist; then, <math>f'(a)</math> exists as well and:
- <math>f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{[h(a)]^2}.</math>
Contents |
Examples
The derivative of <math>(4x - 2)/(x^2 + 1)</math> is:
- <math>\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}</math>
In the example above, the choices
- <math>g(x) = 4x - 2</math>
- <math>h(x) = x^2 + 1</math>
were made. Analogously, the derivative of <math>\sin(x)/x^2</math> (when <math>x</math> ≠ 0) is:
- <math>\frac{\cos(x) x^2 - \sin(x)2x}{x^4}</math>
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(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}</math>
This can be checked by using laws of exponents and the power rule:
- <math>f(x) = \frac{2x^2}{x^3} = \frac{2}{x} = 2x^{-1}</math>
- <math>f'(x) = -2x^{-2} = -\frac{2}{x^2}</math>
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:
<math>f(x) = \frac{|x|+1}{|x|+1},</math>
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.
Proofs
From Newton's difference quotient
Suppose <math>f(x) = g(x)/h(x)</math> where <math>h(x) \neq 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>
We pull out the <math>1/\Delta x</math> and combine the fractions in the numerator:
- <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>
Adding and subtracting <math>g(x)h(x)</math> in the numerator:
- <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>
We factor this and multiply the <math>1/\Delta x</math> through the numerator:
- <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>
Now we move the limit through:
- <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>
By the definition of the difference quotient, the limits in the numerator are derivatives, so we have:
- <math>= \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}</math>
Using the Chain Rule
Consider the identity
- <math> \frac{u}{v}\; =\; \frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]</math>
Then
- <math>\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]</math>
Leading to
- <math>\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]</math>
Multiplying out leads to
- <math>\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]</math>
Finally, taking a common denominator leaves us with the expected result
- <math>\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{\left[ v\frac{du}{dx}-u\frac{dv}{dx} \right]}{v^{2}}</math>
By total differentials
An even more elegant proof is a consequence of the law about total differentials, which states that the total differential,
- <math>dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy + \frac{\partial F}{\partial z} dz + ...</math>
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 <math>F = N(x)/D(x)</math>, then it must be true both that
- (*) <math>dF = \frac{\partial F}{\partial x}dx</math>
and that
- <math>dF = \frac{\partial F}{\partial N}dN + \frac{\partial F}{\partial D}dD</math>.
But we know that <math>dN = N'(x) dx</math> and <math>dD = D'(x) dx</math>.
Substituting and setting these two total differentials equal to one another (since they represent limits which we can manipulate), we obtain the equation
- <math>\frac{\partial F}{\partial x} dx = \frac{\partial F}{\partial N}N'(x) dx + \frac{\partial F}{\partial D}D'(x) dx</math>
which requires that
- (#) <math>\frac{\partial F}{\partial x} = \frac{\partial F}{\partial N}N'(x) + \frac{\partial F}{\partial D}D'(x)</math>.
We compute the partials on the right:
- <math>\frac{\partial F}{\partial N} = \frac{\partial (N/D)}{\partial N} = \frac{1}{D}</math>;
- <math>\frac{\partial F}{\partial D} = \frac{\partial (N/D)}{\partial D} = -\frac{N}{D^2}</math>.
If we substitute them into (#),
- <math>\frac{\partial F}{\partial x} = \frac{N'(x)}{D(x)} - \frac{N(x) D'(x)}{D(x)^2}</math>
- <math>\frac{\partial F}{\partial x} = \frac{D(x)N'(x)}{D(x)^2} - \frac{N(x) D'(x)}{D(x)^2}</math>
which gives us the quotient rule, since, by (*),
- <math>\frac{dF}{dx} = \frac{\partial F}{\partial x}</math>.
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.
See also
bs:Pravilo derivacije količnika ca:Regla del quocient de:Quotientenregel eo:Derivaĵo de kvociento es:Regla del cociente id:Kaidah hasil-bagi is:Brotaregla ko:몫의 규칙 it:Regola del quoziente nl:Quotiëntregel th:กฎผลหาร tr:Bölme kuralı zh:除法定则
