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>\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)</math> ≠ <math>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>
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Examples
The derivative of <math>(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>\sin(x)/x^2</math> (when <math>x</math> ≠ 0) 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)\,</math> | <math>=\frac {\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \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
- Suppose <math>f(x) = g(x)/h(x)</math>
- where <math>h(x)</math>≠ 0 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
- Suppose <math>f(x) = 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>
By total differentials
An even more elegant proof is a consequence of the old 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 so is 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.
Mnemonic
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 D 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 over Ho-Ho."