Quotient rule

In calculus, the quotient rule is a method of finding the derivative of a function (mathematics) 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 :

f(x) = \frac{g(x)}{h(x)}

and

h(x) \ne 0

, then the rule states that the derivative of

g(x)/h(x)

is equal to: :

\frac{d}{dx}f(x) = f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{{h(x)}^2}.

Or more precisely; for all

x

in some open set containing the number

a

, with

h(a) \ne 0

; and, such that

g'(a)

and

h'(a)

both exist; then,

f'(a)

exists as well: :

f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{h(a)^2}

== Examples == The derivative of

\frac{(4x - 2)}{x^2 + 1}

is: :

\frac{d}{dx} \frac{(4x - 2)}{x^2 + 1}

=
\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}

The derivative of

\frac{\sin(x)}{x^2}

(when

x \ne 0

) is: :

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

For more information regarding the derivatives of trigonometric functions, see: derivative. Another example is: :

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

whereas

g(x) = 2x^2

and

h(x) = x^3

, and

g'(x) = 4x

and

h'(x) = 3x^2

. The derivative of

f(x)

is determined as follows: :

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

=
\frac{4x^4 - 6x^4}{x^6}

=
\frac{-2x^4}{x^6}

=
\frac{-2}{x^2}

==Proofs== ===From Newton's difference quotient=== :

\mbox{let }f(x) = \frac{g(x)}{h(x)}

::where

h(x) \ne 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}

= \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]

= \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]

= \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]

= \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)}

= \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))}

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

===From the product rule=== :

\mbox{let }f(x)=\frac{g(x)}{h(x)}

g(x)=f(x)h(x)\mbox{  }

g'(x)=f'(x)h(x) + f(x)h'(x)\mbox{  }

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. :

f'(x)=\frac{g'(x) - f(x)h'(x)}{h(x)} = \frac{g'(x) - \frac{g(x)}{h(x)}\cdot h'(x)}{h(x)}

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

== Mnemonic == It is often memorized as a rhyme type song. "Lo-dee-hi minus hi-dee-lo all over lo-lo"; Lo being the denominator, Hi being the numerator and D being the derivative. ==See also== *Product rule Calculus th:กฎผลหาร

Quotient rule

== Informal proof == Out of curiosity, what exactly is there about the informal proof that makes it informal? User:Cburnett 23:22, 3 Feb 2005 (UTC) :I guess it's considered an informal proof because it's based off the product rule, while the "formal" proof is based off the difference quotient, which is more direct. I changed the headings to be more specific. - User:Evil saltine 05:11, 4 Feb 2005 (UTC)

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