Kvotregeln

Kvotregeln är inom matematisk analys, metoden att finna derivatan till en kvot av två differtierbara funktioner.[1][2][3]

Låt

f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)={\cfrac {g(x)}{h(x)}}}

där

g ( x ) ,   h ( x ) {\displaystyle g(x),\ h(x)}

är differentierbara funktioner och

h ( x ) 0. {\displaystyle h(x)\neq 0.}

Enligt kvotregeln är derivatan av f ( x ) {\displaystyle f(x)}

f ( x ) = g ( x ) h ( x ) g ( x ) h ( x ) [ h ( x ) ] 2 . {\displaystyle f'(x)={\cfrac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}.}

Exempel

d d x e x x 2 = ( d d x e x ) ( x 2 ) ( e x ) ( d d x x 2 ) ( x 2 ) 2 = ( e x ) ( x 2 ) ( e x ) ( 2 x ) x 4 = e x ( x 2 ) x 3 . {\displaystyle {\begin{aligned}{\cfrac {d}{dx}}{\frac {e^{x}}{x^{2}}}&={\cfrac {\left({\cfrac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\cfrac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\cfrac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\cfrac {e^{x}(x-2)}{x^{3}}}.\end{aligned}}}

Kvotregeln kan användas till att finna derivatan av f ( x ) = tan x = sin x cos x   {\displaystyle f(x)=\tan x={\cfrac {\sin x}{\cos x}}\ \Rightarrow }

d d x tan x = d d x sin x cos x = ( d d x sin x ) ( cos x ) ( sin x ) ( d d x cos x ) cos 2 x = cos 2 x + sin 2 x cos 2 x = 1 cos 2 x = sec 2 x . {\displaystyle {\begin{aligned}{\cfrac {d}{dx}}\tan x&={\cfrac {d}{dx}}{\cfrac {\sin x}{\cos x}}\\&={\cfrac {\left({\cfrac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\cfrac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\cfrac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\cfrac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}}

Bevis

Låt

f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)={\cfrac {g(x)}{h(x)}}}

Tillämpa definitionerna av derivator och egenskaperna hos gränsvärden:

f ( x ) = lim k 0 f ( x + k ) f ( x ) k = lim k 0 g ( x + k ) h ( x + k ) g ( x ) h ( x ) k = lim k 0 g ( x + k ) h ( x ) g ( x ) h ( x + k ) k h ( x ) h ( x + k ) = lim k 0 g ( x + k ) h ( x ) g ( x ) h ( x + k ) k lim k 0 1 h ( x ) h ( x + k ) = ( lim k 0 g ( x + k ) h ( x ) g ( x ) h ( x ) + g ( x ) h ( x ) g ( x ) h ( x + k ) k ) 1 h ( x ) 2 = ( lim k 0 g ( x + k ) h ( x ) g ( x ) h ( x ) k lim k 0 g ( x ) h ( x + k ) g ( x ) h ( x ) k ) 1 h ( x ) 2 = ( h ( x ) lim k 0 g ( x + k ) g ( x ) k g ( x ) lim k 0 h ( x + k ) h ( x ) k ) 1 h ( x ) 2 = g ( x ) h ( x ) g ( x ) h ( x ) h ( x ) 2 . {\displaystyle {\begin{aligned}f'(x)&=\lim _{k\to 0}{\cfrac {f(x+k)-f(x)}{k}}\\&=\lim _{k\to 0}{\cfrac {{\cfrac {g(x+k)}{h(x+k)}}-{\cfrac {g(x)}{h(x)}}}{k}}\\&=\lim _{k\to 0}{\cfrac {g(x+k)h(x)-g(x)h(x+k)}{k\cdot h(x)h(x+k)}}\\&=\lim _{k\to 0}{\cfrac {g(x+k)h(x)-g(x)h(x+k)}{k}}\cdot \lim _{k\to 0}{\cfrac {1}{h(x)h(x+k)}}\\&=\left(\lim _{k\to 0}{\cfrac {g(x+k)h(x)-g(x)h(x)+g(x)h(x)-g(x)h(x+k)}{k}}\right)\cdot {\cfrac {1}{h(x)^{2}}}\\&=\left(\lim _{k\to 0}{\cfrac {g(x+k)h(x)-g(x)h(x)}{k}}-\lim _{k\to 0}{\cfrac {g(x)h(x+k)-g(x)h(x)}{k}}\right)\cdot {\cfrac {1}{h(x)^{2}}}\\&=\left(h(x)\lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}-g(x)\lim _{k\to 0}{\cfrac {h(x+k)-h(x)}{k}}\right)\cdot {\cfrac {1}{h(x)^{2}}}\\&={\cfrac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}

Se även

  • Produktregeln
  • Partialintegration

Referenser

Noter

  1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th). Brooks/Cole. ISBN 0-495-01166-5. https://archive.org/details/calculusearlytra00stew_1 
  2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th). Brooks/Cole. ISBN 0-547-16702-4 
  3. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th). Addison-Wesley. ISBN 0-321-58876-2