Encyclopedia > Quotient rule

  Article Content

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>

All Wikipedia text is available under the terms of the GNU Free Documentation License

  Search Encyclopedia

Search over one million articles, find something about almost anything!
  Featured Article
Sanskrit language

... in Latin and Greek combined. The language underwent several stages of consolidation and modification. In its older Vedic form, it is a close descendant of ...

This page was created in 101.8 ms