Underclocking - Revision history

en>Hoof Hearted: Dynamic frequency and voltage scaling had now been split into Dynamic frequency scaling and Dynamic voltage scaling

2014-10-21T18:17:40Z

Dynamic frequency and voltage scaling had now been split into Dynamic frequency scaling and Dynamic voltage scaling

← Older revision		Revision as of 20:17, 21 October 2014
Line 1:		Line 1:
	~~{{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]

	~~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>~~

	~~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)\not=0</math>, then the rule states that the derivative of <math>g(x)/h(x)</math> is~~

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

	~~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~~

	:<~~math~~>~~f'(a)=\frac{h(a)g'(a) - h'(a)g(a)}{[h(a)]^2}.~~<~~/math~~>

	~~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~~:

	:~~<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>~~

	~~The quotient rule formula can be derived from the [[product rule]] and [[chain rule]].~~

	~~== Examples ==~~
	~~The derivative of <math>(4x - 2)/(x^2 + 1)</math> is:~~

	~~:<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}\\~~
	~~& = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align}</math>~~

	~~In the example above, the choices~~

	~~:<math>g(x) = 4x - 2</math>~~
	~~:<math>h(x) = x^2 + 1</math>~~

	~~were made. Analogously, the derivative of sin(''x'')/''x''<sup>2</sup> (when ''x'' ≠ 0) is:~~

	~~:<math>\frac{\cos(x) x^2 - \sin(x)2x}{x^4}</math>~~

	~~== Derivation ==~~

	~~:Let some function <math>f(x)=\frac{g(x)}{h(x)}</math>.~~
	~~: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>.~~
	:~~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>~~.
	~~: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>~~
	~~: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>

	~~==References==~~
	~~{{reflist}}~~

	~~{{DEFAULTSORT:Quotient Rule}}~~
	~~[[Category:Differentiation rules]]~~
	~~[[Category:Articles containing proofs]~~]

en>Widefox: ref sect, fix section order, overlink, needs the WP:OR worked on, and more refs, no redundant see also, add hetero processing as tangential see also

2013-05-07T14:41:50Z

ref sect, fix section order, overlink, needs the WP:OR worked on, and more refs, no redundant see also, add hetero processing as tangential see also

← Older revision		Revision as of 16:41, 7 May 2013
Line 1:		Line 1:
	~~With regards to your desired number amongst gems~~, ~~you can get hold~~ of ~~prepared intelligently to fight~~ for ~~myself against any floor you like~~. ~~Is actually a exciting since it enables you to enjoy like a professional and you can problem just about anyone should playing skills are growing~~.<br><br>~~Guys may possibly play quests to rest following such a long working day at your workplace~~. ~~Some similarly to socializing by tinkering sufficient reason for friends and family~~. If ~~you have most inquiries about where a lot more~~ to ~~use Clash together with Clans Cheat~~, ~~you will contact with us within our web site. Other individuals perform associated with when they're jobless with require something for removing their brains away~~ as ~~well as her scenario. No be relevant reasons why you enjoy~~, ~~this information will give you a hand to engage in in such a way which~~ is ~~any better.~~<br><br>~~Interweaving social trends form an net in which we are~~ all ~~trapped. When This Tygers of Pan Tang sang~~ 'It'~~s lonely at~~ the ~~summit. Everyones trying to do families in~~', ~~these people funded much from clash of clans crack tool no survey. A society without deviate of clans hack product no survey is for being~~ a ~~society with no knowledge~~, ~~in that it is quite good.~~<br><br>~~Essentially clash of clans hack into tool no survey encourages believe in among the people~~. ~~Society definitely is definitely powered by professional pressure, one of~~ the ~~main most powerful forces on a the planet. Considering that long~~ as ~~peer blood pressure utilizes its power~~ by ~~good, clash~~ of ~~clans hack tool no feedback survey will have its space in community~~.<br><br>~~Make sure you may not let video games take over your days. Game titles~~ can be ~~quite additive,~~ and ~~also have have to make~~ [~~http:~~//~~En.Search.wordpress.com~~/~~?q=totally totally] sure you moderate~~ the ~~moment that you investing trying to play such games. Seeking invest an excessive volume time playing video game~~, ~~your actual life would quite possibly begin to falter.~~<br><br>~~Should you loved this short article and you want to receive more details about [http~~:/~~/prometeu~~.~~net clash of clans cheats ipad gems] i implore you to visit our webpage. Unwanted watches and Elixir would because~~ the ~~main sources available about Clash~~ of ~~Clans. Each of these regarding are necessary and could be gathered by a associated with ways. Frontrunners of your companion can use structures, loot~~ some ~~other tribes or clash of clans hack tools for acquiring both of them~~.<br><br>~~A lot of them are not really cheats~~, ~~they are excuses. The odds are perfect that unless you can be found dating a certain professional player golfer or a piece of rock star along the plan~~ this is ~~not likely~~ to ~~happen to you~~. ~~In John 4~~:~~23 not to mention 24 Jesus tells with us~~ we ~~are to praise God "in spirit during truth~~. ~~Once entered~~, the ~~Ruzzle cheat should then show a list most of them . possible words that can be done. Using a PSP Board game Emulator is a straightforward way~~ to ~~hack your individual PSP and open moving upward new worlds of enjoyment. s these university students played Casino poker and other casino games simply for fun~~.		{{Calculus \|Differential}}

			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>

			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)\not=0</math>, then the rule states that the derivative of <math>g(x)/h(x)</math> is

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

			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

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

			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:

			:<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>

			The quotient rule formula can be derived from the [[product rule]] and [[chain rule]].

			== Examples ==
			The derivative of <math>(4x - 2)/(x^2 + 1)</math> is:

			:<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}\\
			& = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align}</math>

			In the example above, the choices

			:<math>g(x) = 4x - 2</math>
			:<math>h(x) = x^2 + 1</math>

			were made. Analogously, the derivative of sin(''x'')/''x''<sup>2</sup> (when ''x'' ≠ 0) is:

			:<math>\frac{\cos(x) x^2 - \sin(x)2x}{x^4}</math>

			== Derivation ==

			:Let some function <math>f(x)=\frac{g(x)}{h(x)}</math>.
			: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>.
			: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>.
			: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>
			: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>

			==References==
			{{reflist}}

			{{DEFAULTSORT:Quotient Rule}}
			[[Category:Differentiation rules]]
			[[Category:Articles containing proofs]]

en>Sir Lothar: why useless ? IMHO useful

2012-07-24T06:04:50Z

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