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| {{Calculus |Differential}}
| | My name: Ashlee Rebell<br>Age: 18<br>Country: Switzerland<br>Home town: Schmiedrued <br>ZIP: 5046<br>Address: Pfaffacherweg 16<br><br>Feel free to visit my weblog :: [http://hemorrhoidtreatmentfix.com/prolapsed-hemorrhoid-treatment prolapsed hemorrhoids treatment] |
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| In [[calculus]], the '''quotient rule''' is a method of finding the [[derivative]] of a [[function (mathematics)|function]] that is the [[quotient]] of two other functions for which derivatives exist.<ref>{{cite book | last=Stewart | first=James | authorlink=James Stewart (mathematician) | title=Calculus: Early Transcendentals |publisher=[[Brooks/Cole]] | edition=6th | year=2008 | isbn=0-495-01166-5}}</ref><ref>{{cite book | last1=Larson | first1=Ron | authorlink=Ron Larson (mathematician)| last2=Edwards | first2=Bruce H. | title=Calculus | publisher=[[Brooks/Cole]] | edition=9th | year=2009 | isbn=0-547-16702-4}}</ref><ref>{{cite book | last1 = Thomas | first1 = George B. | last2=Weir | first2= Maurice D. | last3=Hass | first3=Joel | author3-link = Joel Hass | authorlink=George B. Thomas | title=Thomas' Calculus: Early Transcendentals | publisher=[[Addison-Wesley]] | year=2010 | edition=12th | isbn=0-321-58876-2}}</ref>
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| If the function one wishes to differentiate, <math>f(x)</math>, can be written as
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| :<math>f(x) = \frac{g(x)}{h(x)}</math> | |
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| and <math>h(x)\not=0</math>, then the rule states that the derivative of <math>g(x)/h(x)</math> is
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| :<math>f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.</math> | |
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| More precisely, if all ''x'' in some [[open set]] containing the number ''a'' satisfy <math>h(x)\not=0</math>, and <math>g'(a)</math> and <math>h'(a)</math> both exist, then <math>f'(a)</math> exists as well and
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| :<math>f'(a)=\frac{h(a)g'(a) - h'(a)g(a)}{[h(a)]^2}.</math> | |
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| And this can be extended to calculate the second derivative as well (you can prove this by taking the derivative of <math>f(x)=g(x)(h(x))^{-1}</math> twice). The result of this is:
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| :<math>f''(x)=\frac{g''(x)[h(x)]^2-2g'(x)h(x)h'(x)+g(x)[2[h'(x)]^2-h(x)h''(x)]}{[h(x)]^3}.</math> | |
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| The quotient rule formula can be derived from the [[product rule]] and [[chain rule]].
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| == Examples ==
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| The derivative of <math>(4x - 2)/(x^2 + 1)</math> is:
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| :<math>\begin{align}\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(4)(x^2 + 1) - (4x - 2)(2x)}{(x^2 + 1)^2}\\
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| & = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align}</math>
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| In the example above, the choices
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| :<math>g(x) = 4x - 2</math>
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| :<math>h(x) = x^2 + 1</math>
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| were made. Analogously, the derivative of sin(''x'')/''x''<sup>2</sup> (when ''x'' ≠ 0) is:
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| :<math>\frac{\cos(x) x^2 - \sin(x)2x}{x^4}</math>
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| == Derivation ==
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| :Let some function <math>f(x)=\frac{g(x)}{h(x)}</math>.
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| :We wish to find <math>\frac{d}{dx}f(x)</math>, and this is equivalent to <math>\frac{d}{dx}\frac{g(x)}{h(x)}</math>.
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| :We also know that <math>g(x)=f(x)\cdot h(x)</math>. | |
| :By the product rule, we can say that <math>\frac{d}{dx}g(x)=h(x)\cdot\frac{df(x)}{dx}+\frac{dh(x)}{dx}\cdot f(x)</math>.
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| :From the equation above, <math>\frac{d}{dx}f(x)=\frac{\frac{dg(x)}{dx}-\frac{dh(x)}{dx}\cdot f(x)}{h(x)}</math>
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| :Because <math>f(x)=\frac{g(x)}{h(x)}</math>, the right side simplifies to <math>\frac{\frac{dg(x)}{dx}-\frac{dh(x)}{dx}\cdot\frac{g(x)}{h(x)}}{h(x)}</math>, which simplifies to <math>\frac{d}{dx}f(x)=\frac{\frac{dg(x)}{dx}\cdot h(x)-\frac{dh(x)}{dx}\cdot g(x)}{h^2(x)}</math>, completing the derivation.<ref>{{cite book | last=Berresford | first=Geoffrey | last2=Rockett| first2=Andrew| title=Brief Applied Calculus | edition=5th | year=2008 | isbn=0-547-16977-9}}</ref>
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| ==References==
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| {{reflist}}
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| {{DEFAULTSORT:Quotient Rule}}
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| [[Category:Differentiation rules]]
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| [[Category:Articles containing proofs]]
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My name: Ashlee Rebell
Age: 18
Country: Switzerland
Home town: Schmiedrued
ZIP: 5046
Address: Pfaffacherweg 16
Feel free to visit my weblog :: prolapsed hemorrhoids treatment