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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, f(x), can be written as

<math>f(x) = \frac{g(x)}{h(x)}</math>

and h(x) ≠ 0; then, the rule states that the derivative of g(x) / h(x) is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator:

<math>f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{{h(x)}^2}.</math>

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:

<math>f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{h(a)^2}</math>


The derivative of (4x - 2) / (x2 + 1) = [(x2 + 1)(4) - (4x - 2)(2x)] / (x2 + 1)2 = [(4x2 + 4) - (8x2 - 4x)] / (x2 + 1)2 = [-4x2 + 4x + 4] / (x2 + 1)2

The derivative of [sin(x)] / x2 (when x ≠ 0) is ([cos(x)]x2 - [sin(x)](2x)) / x4. For more information regarding the derivatives of trigonometric functions, see: derivative.

Informal Proof A proof of this rule can be derived from Newton's difference quotient: The derivative of [f(x)] / [g(x)] = (the limit as h approaches 0):

<math>\frac{1}{\Delta x} \left(\frac{g(x+\Delta x)}{h(x+\Delta x)} - \frac{g(x)}{h(x)}\right) </math>
<math>= \frac{1}{\Delta x} \frac{g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)}</math> having multiplied the fraction by: g(x)(x + Δx) / g(x)(x + Δx)
<math>= \frac{1}{\Delta x} \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)}</math> adding and subtracting the same value to alllow for an algebraic manipulation
<math>= \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> seperating into groups with common multiples
which for Δx → 0 converges to
<math>\frac{g'(x)h(x) - g(x)h'(x)}{h^2(x)}</math>

To turn this into a proper proof, one has to pick Δx so small that the denominators are all non-zero (and one has to argue that this is always possible because the involved functions are continuous).

Alternate Informal Proof

Using only the product rule:


<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 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.

<math>f'(x)=\frac{g'(x) - f(x)h'(x)}{h(x)}</math>

<math>f'(x)=\frac{g'(x)h(x) - (f(x)h(x))h'(x)}{\left(h(x)\right)^2}</math>

<math>f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{\left(h(x)\right)^2}</math>

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