March 24, 2013, at 02:12 AM by 31.25.19.69 -

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Endelig følger det af, at {$g(x)$} er differentiabel og dermed kontinuert, at {$g(x+\Delta x)$} går mod {$g(x)$}, når {$\Delta x$} går mod 0.

Vi kan hermed konkludere {$$\left( \frac{f(x)}{g(x)}\right)'= \lim_{\Delta x\rightarrow 0}\left(a_s \right) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$$}

March 24, 2013, at 01:54 AM by 31.25.19.69 -

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Brøken {$\displaystyle \frac{f(x+ \Delta x) - f(x)}{\Delta x}$} er differenskvotienten for funktionen f

to:

Brøken {$\displaystyle \frac{f(x+ \Delta x) - f(x)}{\Delta x}$} er differenskvotienten for funktionen f og går derfor mod {$f'(x)$}, når {$\Delta x$} går mod 0

Brøken {$\displaystyle \frac{g(x+ \Delta x) - g(x)}{\Delta x}$} er differenskvotienten for funktionen g og går derfor mod {$g'(x)$}, når {$\Delta x$} går mod 0

March 24, 2013, at 01:52 AM by 31.25.19.69 -

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test {$$f(x)$$}

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Brøken {$\displaystyle \frac{f(x+ \Delta x) - f(x)}{\Delta x}$} er differenskvotienten for funktionen f

March 24, 2013, at 01:49 AM by 31.25.19.69 -

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Brøken foroven splittes op i to

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Endelig splittes brøken foroven op i to

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Vi ser nu på, hvad der sker, når {$\Delta x$} går mod 0.

test {$$f(x)$$}

March 24, 2013, at 01:39 AM by 31.25.19.69 -

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1. Først opskrives differenskvotienten for funktionen {$\displaystyle \frac{f(x)}{g(x)}$}

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Trin 1: Differenskvotienten for funktionen {$\displaystyle \frac{f(x)}{g(x)}$} opskrives

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2. Derefter regner vi lidt på den. Først samles de to brøker i tælleren på en fælles brøkstreg

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Trin 2: Differenskvotienten omskrives. Først samles de to brøker i tælleren på en fælles brøkstreg

March 24, 2013, at 01:37 AM by 31.25.19.69 -

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March 24, 2013, at 01:37 AM by 31.25.19.69 -

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En brøk {$\frac{f(x)}{g(x)}$} hvor {$f(x)$} og {$g(x)$} er differentiable funktioner, differentieres på følgende måde:

to:

En brøk {$\displaystyle \frac{f(x)}{g(x)}$} hvor {$f(x)$} og {$g(x)$} er differentiable funktioner, differentieres på følgende måde:

March 24, 2013, at 01:36 AM by 31.25.19.69 -

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En brøk {$\frac{f(x)}{g(x)}$} hvor {$f(x)$} og {$g(x)$} er differentiable funktioner, differentieres på følgende måde:

March 24, 2013, at 01:31 AM by 31.25.19.69 -

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Der sættes uden for parentes

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g(x) og f(x) sættes uden for parentes

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March 24, 2013, at 01:20 AM by 31.25.19.69 -

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{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)+f(x)\cdot g(x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x)-f(x)\cdot g(x+\Delta x)+f(x)\cdot g(x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

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{$$a_s=\frac{\displaystyle \frac{(f(x+\Delta x)-f(x))\cdot g(x)-f(x)\cdot ((g(x+\Delta x)-g(x))}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

{$$a_s=\frac{\displaystyle \frac{(f(x+\Delta x)-f(x))\cdot g(x)-f(x)\cdot (g(x+\Delta x)-g(x))}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

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{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)-f(x))}{\Delta x}\cdot g(x)-f(x)\cdot \frac{(g(x+\Delta x)-g(x))}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)-f(x))}{\Delta x}\cdot g(x)-f(x)\cdot \frac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 24, 2013, at 01:18 AM by 31.25.19.69 -

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{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}

{$\Delta x$} og {$g(x)\cdot (g(x+\Delta x)$} bytter plads

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x+\Delta x) }{g(x)\cdot g(x+\Delta x)}}{\Delta x}$$}

{$\Delta x$} og {$g(x)\cdot g(x+\Delta x)$} bytter plads

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x+\Delta x)}{\Delta x}}{g(x)\cdot g(x+\Delta x)}$$}

March 23, 2013, at 10:05 PM by 31.25.19.69 -

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Brøken foroven splittes op i to

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)-f(x))}{\Delta x}\cdot g(x)-f(x)\cdot \frac{(g(x+\Delta x)-g(x))}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 23, 2013, at 09:56 PM by 31.25.19.69 -

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Der sættes uden for parentes

{$$a_s=\frac{\displaystyle \frac{(f(x+\Delta x)-f(x))\cdot g(x)-f(x)\cdot ((g(x+\Delta x)-g(x))}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 23, 2013, at 09:53 PM by 31.25.19.69 -

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Vi trækker {$f(x)\cdot g(x)$} fra og lægger det til igen (i tælleren)

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)+f(x)\cdot g(x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 22, 2013, at 12:45 PM by 31.25.19.69 -

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Brøken foroven splittes op i to

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)}{\Delta x}-\frac{f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

March 22, 2013, at 12:44 PM by 31.25.19.69 -

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Brøken foroven splittes op i to

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)}{\Delta x}-\frac{f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 22, 2013, at 12:41 PM by 31.25.19.69 -

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{$$\displaystyle a_s=\frac{\displaystyle \frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

to:

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

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{$$\displaystyle a_s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}

to:

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}

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{$$\displaystyle a_s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

{$$a_s=\frac{\displaystyle \frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 22, 2013, at 12:40 PM by 31.25.19.69 -

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{$$\displaystyle a_s=\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

to:

{$$\displaystyle a_s=\frac{\displaystyle \frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

March 22, 2013, at 12:40 PM by 31.25.19.69 -

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{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}

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{$$\displaystyle a_s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}

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{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

to:

{$$\displaystyle a_s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 22, 2013, at 12:39 PM by 31.25.19.69 -

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1. Først opskrives differenskvotienten for funktionen {$\frac{f(x)}{g(x)}$}

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1. Først opskrives differenskvotienten for funktionen {$\displaystyle \frac{f(x)}{g(x)}$}

March 22, 2013, at 12:38 PM by 31.25.19.69 -

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{$$a_s=\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

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{$$\displaystyle a_s=\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

March 22, 2013, at 12:37 PM by 31.25.19.69 -

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{$Delta x$} flyttes op i nævneren på den øverste brøk

{$$s=\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x) \cdot \Delta x}$$}

to:

{$\Delta x$} og {$g(x)\cdot (g(x+\Delta x)$} bytter plads

{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x)}{\Delta x}}{g(x)\cdot (g(x+\Delta x)}$$}

March 22, 2013, at 12:28 PM by 31.25.19.69 -

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{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot \cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}}

to:

{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}

{$Delta x$} flyttes op i nævneren på den øverste brøk

{$$s=\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x) \cdot \Delta x}$$}

March 22, 2013, at 12:22 PM by 31.25.19.69 -

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{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)\cdot (g(x+\Delta x)-f(x)\cdot }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}}

to:

{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot \cdot (g(x+\Delta x) }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}}

March 22, 2013, at 12:22 PM by 31.25.19.69 -

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2. Derefter regner vi lidt på den. Først samles de to brøker i tælleren på en fælles brøkstreg

{$$s=\frac{\frac{f(x+\Delta x)\cdot g(x)\cdot (g(x+\Delta x)-f(x)\cdot }{g(x)\cdot (g(x+\Delta x)}}{\Delta x}$$}}

March 18, 2013, at 12:14 AM by 31.25.19.69 -

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1. {$$a_s=\frac{\frac{f(x+\Delta x}{1}}{\Delta x}$$}

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1. Først opskrives differenskvotienten for funktionen {$\frac{f(x)}{g(x)}$}

{$$a_s=\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}$$}

March 18, 2013, at 12:10 AM by 31.25.19.69 -

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1. {$$a_s=\frac{\frac{f(x+\Delta x}{1}}{\Delta x}$$}

March 18, 2013, at 12:08 AM by 31.25.19.69 -

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Bevis:

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Bevis:

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(:toggle div=brøkbevis init=show button=1 lshow=Bevis lhide="Skjul Bevis":)

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(:toggle div=broekbevis init=show button=1 lshow=Bevis lhide="Skjul Bevis":)

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(:toggle div=brøkbevis init=show button=1 lshow=Bevis lhide="Skjul Bevis":)

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(:table border=1 width=60% cellpadding=10 align=center bgcolor=#cccc99 cellspacing=0 :)

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(:table border=1 width=40% cellpadding=10 align=center bgcolor=#cccc99 cellspacing=0 :)

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Vi bruger tretrinsreglen

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(:table border=1 width=60% cellpadding=10 align=center bgcolor=#cccc99 cellspacing=0 :) (:cellnr:)

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(:tableend:)

March 05, 2013, at 11:28 PM by 31.25.19.69 -

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

to:

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

Bevis:

March 05, 2013, at 11:27 PM by 31.25.19.69 -

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

to:

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

March 05, 2013, at 11:27 PM by 31.25.19.69 -

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

to:

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

March 05, 2013, at 11:26 PM by 31.25.19.69 -

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{$$(\frac{f(x)}{g(x)}'= \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$$}

to:

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

March 05, 2013, at 11:24 PM by 31.25.19.69 -

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

to:

{$$(\frac{f(x)}{g(x)}'= \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$$}

March 05, 2013, at 11:23 PM by 31.25.19.69 -

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{$$(f(g(x))'=f'(g(x))\cdot g'(x)$$}

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

March 05, 2013, at 11:22 PM by 31.25.19.69 -

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(:noleft:) (:noheader:) (:notitle:)

Differentiation af brøk

Formel for differentiation af en brøk:

{$$(f(g(x))'=f'(g(x))\cdot g'(x)$$}

MatematikA.DifferentiationAfBrøk History

Bevis:

Bevis:

Bevis:

Differentiation af brøk