Lösung 1.2:1f - Versionsgeschichte

Sekretariat1 um 07:35, 20. Aug. 2009

Sekretariat1 — Thu, 20 Aug 2009 07:35:35 GMT

← Nächstältere Version		Version vom 07:35, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir verwenden die Quotientenregel,	+	Wir verwenden die Quotientenregel.

-	{{Abgesetzte Formel\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{}</math>}}

-	Den Ausdruck <math>x\ln x</math> können wir mit der Faktorregel ableiten,	+	Den Ausdruck <math>x\ln x</math> können wir mit der Faktorregel ableiten.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 9:		Zeile 9:
	&= (x)'\ln x + x\,(\ln x)'\\[5pt]		&= (x)'\ln x + x\,(\ln x)'\\[5pt]
	&= 1\cdot\ln x + x\cdot\frac{1}{x}\\[5pt]		&= 1\cdot\ln x + x\cdot\frac{1}{x}\\[5pt]
-	&= \ln x+1\,\textrm{.}	+	&= \ln x+1\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

Martin um 19:15, 18. Apr. 2009

Martin — Sat, 18 Apr 2009 19:15:28 GMT

← Nächstältere Version		Version vom 19:15, 18. Apr. 2009
Zeile 1:		Zeile 1:
-	In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,	+	Wir verwenden die Quotientenregel,

	{{Abgesetzte Formel\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}

-	~~We can, in turn, differentiate the expression~~ <math>x\ln x</math> ~~by using the product rule~~,	+	Den Ausdruck <math>x\ln x</math> können wir mit der Faktorregel ableiten,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 12:		Zeile 12:
	\end{align}</math>}}		\end{align}</math>}}

-	~~All in all, we thus obtain~~	+	Wir erhalten so

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

Tekbot: Solution 1.2:1f moved to Lösung 1.2:1f: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:03:15 GMT

Solution 1.2:1f moved to Lösung 1.2:1f: Robot: moved page

← Nächstältere Version

Version vom 10:03, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 12:52:44 GMT

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

← Nächstältere Version		Version vom 12:52, 10. Mär. 2009
Zeile 1:		Zeile 1:
	In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,		In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,

-	{{~~Displayed math~~\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}

	We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,		We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(x\ln x)'		(x\ln x)'
	&= (x)'\ln x + x\,(\ln x)'\\[5pt]		&= (x)'\ln x + x\,(\ln x)'\\[5pt]
Zeile 14:		Zeile 14:
	All in all, we thus obtain		All in all, we thus obtain

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

Tek um 14:22, 14. Okt. 2008

Tek — Tue, 14 Oct 2008 14:22:38 GMT

← Nächstältere Version		Version vom 14:22, 14. Okt. 2008
Zeile 1:		Zeile 1:
-	In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by	+	In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,
-	<math>\sin x</math>	+
-	". As a first step, we therefore use the quotient rule,	+

		+	{{Displayed math\|\|<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}

-	~~<math>\left( \frac{x\ln x}{\sin x} \right)^{\prime }=\frac{\left( x\ln x \right)^{\prime }\centerdot \sin x-x\ln x\centerdot \left( \sin x \right)^{\prime }}{\left( \sin x \right)^{2}}</math>~~	+	We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,
-		+
-		+
-	We can, in turn, differentiate the expression	+
-	<math>x\ln x</math>	+
-	by using the product rule:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \left( x\ln x \right)^{\prime }=\left( x \right)^{\prime }\ln x+x\left( \ln x \right)^{\prime } \\~~	+
-	~~& \\~~	+
-	~~& =1\centerdot \ln x+x\centerdot \frac{1}{x}=\ln x+1 \\~~	+
-	~~\end{align}</math>~~	+

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

	All in all, we thus obtain		All in all, we thus obtain

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

Tekbot: Lösning 1.2:1f moved to Solution 1.2:1f: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:23:41 GMT

Lösning 1.2:1f moved to Solution 1.2:1f: Robot: moved page

← Nächstältere Version

Version vom 14:23, 16. Sep. 2008

Ian um 15:53, 12. Sep. 2008

Ian — Fri, 12 Sep 2008 15:53:27 GMT

← Nächstältere Version		Version vom 15:53, 12. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by
-	~~<center> [[Image:1_2_1f~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>\sin x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	". As a first step, we therefore use the quotient rule,
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:1_2_1f~~-2(2)~~.gif]] </center>~~	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\left( \frac{x\ln x}{\sin x} \right)^{\prime }=\frac{\left( x\ln x \right)^{\prime }\centerdot \sin x-x\ln x\centerdot \left( \sin x \right)^{\prime }}{\left( \sin x \right)^{2}}</math>
		+
		+
		+	We can, in turn, differentiate the expression
		+	<math>x\ln x</math>
		+	by using the product rule:
		+
		+
		+	<math>\begin{align}
		+	& \left( x\ln x \right)^{\prime }=\left( x \right)^{\prime }\ln x+x\left( \ln x \right)^{\prime } \\
		+	& \\
		+	& =1\centerdot \ln x+x\centerdot \frac{1}{x}=\ln x+1 \\
		+	\end{align}</math>
		+
		+
		+	All in all, we thus obtain
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{x\ln x}{\sin x} \right)^{\prime }=\frac{\left( \ln x+1 \right)\centerdot \sin x-x\ln x\centerdot \cos x}{\left( \sin x \right)^{2}} \\
		+	& \\
		+	& =\frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin ^{2}x} \\
		+	\end{align}</math>

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

Tekbot — Thu, 21 Aug 2008 07:51:34 GMT

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

← Nächstältere Version		Version vom 07:51, 21. Aug. 2008
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Oskar — Fri, 04 Apr 2008 08:34:35 GMT

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