Lösung 1.2:4b - Versionsgeschichte

Stefanie um 15:19, 1. Okt. 2009

Stefanie — Thu, 01 Oct 2009 15:19:42 GMT

← Nächstältere Version		Version vom 15:19, 1. Okt. 2009
Zeile 10:		Zeile 10:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	(\sin\ln x + \cos\ln x)'	+	(\sin(\ln x) + \cos(\ln x))'
-	&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]	+	&= (\sin(\ln x))' + (\cos(\ln x))'\\[5pt]
-	&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt]	+	&= \cos(\ln x)\cdot (\ln x)' - \sin(\ln x)\cdot (\ln x)'\\[5pt]
-	&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\,\textrm{.}	+	&= \cos(\ln x)\cdot\frac{1}{x} - \sin(\ln x)\cdot\frac{1}{x}\,\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Zeile 19:		Zeile 19:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]	+	\frac{d}{dx}\,\bigl[x(\sin(\ln x) + \cos(\ln x))\bigr]
-	&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]	+	&= \sin (\ln x) + \cos (\ln x) + \cos (\ln x) - \sin (\ln x)\\[5pt]
-	&= 2\cos \ln x\,\textrm{.}	+	&= 2\cos (\ln x)\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Zeile 27:		Zeile 27:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\frac{d}{dx}\,2\cos\ln x	+	\frac{d}{dx}\,2\cos(\ln x)
-	&= -2\sin\ln x\cdot (\ln x)'\\[5pt]	+	&= -2\sin(\ln x)\cdot (\ln x)'\\[5pt]
-	&= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]	+	&= -2\sin(\ln x)\cdot \frac{1}{x}\\[5pt]
-	&= -\frac{2\sin\ln x}{x}\,\textrm{.}	+	&= -\frac{2\sin(\ln x)}{x}\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Sekretariat1 um 11:09, 20. Aug. 2009

Sekretariat1 — Thu, 20 Aug 2009 11:09:33 GMT

← Nächstältere Version		Version vom 11:09, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir berechnen zuerst die erste Ableitung mit der ~~Fakrorregel,~~	+	Wir berechnen zuerst die erste Ableitung mit der Faktorregel

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	Wir ~~leiten~~ leiten den zweiten Term, Term für Term mit der Kettenregel ab,	+	Wir leiten den zweiten Term, Term für Term mit der Kettenregel ab

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]		&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
	&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt]		&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt]
-	&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.}	+	&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Martin um 12:37, 19. Apr. 2009

Martin — Sun, 19 Apr 2009 12:37:01 GMT

← Nächstältere Version		Version vom 12:37, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~To start with, we determine the first derivative and begin by using the product rule~~,	+	Wir berechnen zuerst die erste Ableitung mit der Fakrorregel,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We divide up the differentiation of the second term in sections and use the chain rule~~,	+	Wir leiten leiten den zweiten Term, Term für Term mit der Kettenregel ab,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 16:		Zeile 16:
	\end{align}</math>}}		\end{align}</math>}}

-	~~This means that~~	+	Also haben wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 24:		Zeile 24:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The second derivative is~~	+	Die zweite Ableitung ist

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot: Solution 1.2:4b moved to Lösung 1.2:4b: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:07:55 GMT

Solution 1.2:4b moved to Lösung 1.2:4b: Robot: moved page

← Nächstältere Version

Version vom 10:07, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Tue, 10 Mar 2009 12:55:04 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 12:55, 10. Mär. 2009
Zeile 1:		Zeile 1:
	To start with, we determine the first derivative and begin by using the product rule,		To start with, we determine the first derivative and begin by using the product rule,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]		\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
	&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]		&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]
Zeile 9:		Zeile 9:
	We divide up the differentiation of the second term in sections and use the chain rule,		We divide up the differentiation of the second term in sections and use the chain rule,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(\sin\ln x + \cos\ln x)'		(\sin\ln x + \cos\ln x)'
	&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]		&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
Zeile 18:		Zeile 18:
	This means that		This means that

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]		\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
	&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]		&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]
Zeile 26:		Zeile 26:
	The second derivative is		The second derivative is

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,2\cos\ln x		\frac{d}{dx}\,2\cos\ln x
	&= -2\sin\ln x\cdot (\ln x)'\\[5pt]		&= -2\sin\ln x\cdot (\ln x)'\\[5pt]

Tek um 13:58, 15. Okt. 2008

Tek — Wed, 15 Oct 2008 13:58:58 GMT

← Nächstältere Version		Version vom 13:58, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	To start with, we determine the first derivative and begin by using the product rule:	+	To start with, we determine the first derivative and begin by using the product rule,

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
		+	&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]
		+	&= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	We divide up the differentiation of the second term in sections and use the chain rule,
-	~~& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\~~	+
-	~~& =\left( x \right)^{\prime }\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\~~	+
-	~~& =1\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	We divide up the differentiation of the second term in sections and use the chain rule:	+
-		+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \left( \sin \ln x+\cos \ln x \right)^{\prime }=\left( \sin \ln x \right)^{\prime }+\left( \cos \ln x \right)^{\prime } \\~~	+
-	~~& =\cos \ln x\centerdot \left( \ln x \right)^{\prime }-\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\~~	+
-	~~& =\cos \ln x\centerdot \frac{1}{x}-\sin \ln x\centerdot \frac{1}{x} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	(\sin\ln x + \cos\ln x)'
		+	&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
		+	&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt]
		+	&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.}
		+	\end{align}</math>}}

	This means that		This means that

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
-	& \frac{d}{dx}x\~~left~~( \sin \ln x+\cos \ln x ~~\right~~) \\	+	&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]
-	& =\sin \ln x+\cos \ln x+\cos \ln x-\sin \ln x \\	+	&= 2\cos \ln x\,\textrm{.}
-	& =2\cos \ln x \\	+	\end{align}</math>}}
-	\end{align}</math>	+
-		+

	The second derivative is		The second derivative is

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{d}{dx}\,2\cos\ln x
-	& \frac{d}{dx}2\cos \ln x=-2\sin \ln x\~~centerdot \left~~( \ln x ~~\right~~)~~^{\prime }~~ \\	+	&= -2\sin\ln x\cdot (\ln x)'\\[5pt]
-	& =-2\sin \ln x\~~centerdot~~ \frac{1}{x}=-\frac{2\sin \ln x}{x} \\	+	&= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]
-	\end{align}</math>	+	&= -\frac{2\sin\ln x}{x}\,\textrm{.}
		+	\end{align}</math>}}

Ian um 14:15, 12. Okt. 2008

Ian — Sun, 12 Oct 2008 14:15:48 GMT

← Nächstältere Version		Version vom 14:15, 12. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	To start with, we determine the first derivative and begin by using the product rule:
-	~~<center> [[Image:1_2_4b-~~1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\begin{align}
-	<~~center~~> ~~[[Image:1_2_4b-2~~(2)~~.gif]]~~ </~~center~~>	+	& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\
-	{{~~NAVCONTENT_STOP~~}}	+	& =\left( x \right)^{\prime }\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\
		+	& =1\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\
		+	\end{align}</math>
		+
		+
		+	We divide up the differentiation of the second term in sections and use the chain rule:
		+
		+
		+
		+	<math>\begin{align}
		+	& \left( \sin \ln x+\cos \ln x \right)^{\prime }=\left( \sin \ln x \right)^{\prime }+\left( \cos \ln x \right)^{\prime } \\
		+	& =\cos \ln x\centerdot \left( \ln x \right)^{\prime }-\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\
		+	& =\cos \ln x\centerdot \frac{1}{x}-\sin \ln x\centerdot \frac{1}{x} \\
		+	\end{align}</math>
		+
		+
		+	This means that
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\
		+	& =\sin \ln x+\cos \ln x+\cos \ln x-\sin \ln x \\
		+	& =2\cos \ln x \\
		+	\end{align}</math>
		+
		+
		+	The second derivative is
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}2\cos \ln x=-2\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\
		+	& =-2\sin \ln x\centerdot \frac{1}{x}=-\frac{2\sin \ln x}{x} \\
		+	\end{align}</math>

Tekbot: Lösning 1.2:4b moved to Solution 1.2:4b: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:28:21 GMT

Lösning 1.2:4b moved to Solution 1.2:4b: Robot: moved page

← Nächstältere Version

Version vom 14:28, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:53:54 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

← Nächstältere Version		Version vom 07:53, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_4b-1(2).gif]] </center>	+	<center> [[Image:1_2_4b-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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-	<center> [[~~Bild~~:1_2_4b-2(2).gif]] </center>	+	<center> [[Image:1_2_4b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_4b-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:1_2_4b-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Fri, 04 Apr 2008 08:44:42 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_4b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_4b-2(2).gif </center> {{NAVCONTENT_STOP}}

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