Lösung 1.2:3e - Versionsgeschichte

Sekretariat1 um 10:57, 20. Aug. 2009

2009-08-20T10:57:16Z

← Nächstältere Version		Version vom 10:57, 20. Aug. 2009
Zeile 4:		Zeile 4:
	e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}		e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}

-	Danach verwenden wir nochmals die Kettenregel,	+	Danach verwenden wir nochmals die Kettenregel.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
	e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'		e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'
	&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]		&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]
-	&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.}	+	&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

Sekretariat1 um 10:56, 20. Aug. 2009

2009-08-20T10:56:48Z

← Nächstältere Version		Version vom 10:56, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Die ~~üßerste~~ Funktion ist die Exponentialfunktion. Die Kettenregel ergibt	+	Die äußerste Funktion ist die Exponentialfunktion. Die Kettenregel ergibt

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =

Martin um 12:28, 19. Apr. 2009

2009-04-19T12:28:22Z

← Nächstältere Version		Version vom 12:28, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~At first sight, the expression looks like “''e'' raised to something” and therefore we differentiate using the chain rule,~~	+	Die üßerste Funktion ist die Exponentialfunktion. Die Kettenregel ergibt

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =
	e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}		e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}

-	~~Then, we differentiate “sine of something”~~,	+	Danach verwenden wir nochmals die Kettenregel,

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

Tekbot: Solution 1.2:3e moved to Lösung 1.2:3e: Robot: moved page

2009-03-11T10:06:55Z

Solution 1.2:3e moved to Lösung 1.2:3e: Robot: moved page

← Nächstältere Version

Version vom 10:06, 11. Mär. 2009

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

2009-03-10T12:54:34Z

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

← Nächstältere Version		Version vom 12:54, 10. Mär. 2009
Zeile 1:		Zeile 1:
	At first sight, the expression looks like “''e'' raised to something” and therefore we differentiate using the chain rule,		At first sight, the expression looks like “''e'' raised to something” and therefore we differentiate using the chain rule,

-	{{~~Displayed math~~\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =	+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =
	e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}		e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}

	Then, we differentiate “sine of something”,		Then, we differentiate “sine of something”,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'		e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'
	&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]		&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]
	&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.}		&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 13:12, 15. Okt. 2008

2008-10-15T13:12:24Z

← Nächstältere Version		Version vom 13:12, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	At first sight, the expression looks like “	+	At first sight, the expression looks like “''e'' raised to something” and therefore we differentiate using the chain rule,
-	~~<math>~~e~~</math>~~	+
-	raised to something” and therefore we differentiate using the chain rule:	+

		+	{{Displayed math\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =
		+	e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}

-	~~<math>\frac{d}{dx}e^{\left\{ \left. \sin x^{2} \right\} \right.}=e^{\left\{ \left. \sin x^{2} \right\} \right.}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }</math>~~	+	Then, we differentiate “sine of something”,

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~Then, we differentiate “sine of something”:~~	+	e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'
-		+	&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]
-		+	&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.}
-	<math>\begin{align}	+	\end{align}</math>}}
-	& e^{\sin x^{2}}\~~centerdot~~ \~~left~~( \~~left~~\{ ~~\left. \sin~~ x^{2} \~~right\} \right. \right~~)~~^{\prime }~~=e^{\sin x^{2}}\~~centerdot~~ \cos \~~left\~~{ ~~\left.~~ x^{2} \~~right~~\} \~~right~~.~~\centerdot \left( \left\~~{ ~~\left.~~ x^{2} \~~right\} \right. \right~~)~~^{\prime }~~ \\	+
-	& =e^{\sin x^{2}}\~~centerdot~~ \cos x^{2}\~~centerdot~~ 2x \\	+
-	\end{align}</math>	+

Ian um 13:27, 12. Okt. 2008

2008-10-12T13:27:58Z

← Nächstältere Version		Version vom 13:27, 12. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	At first sight, the expression looks like “
-	<~~center~~> ~~[[Image~~:~~1_2_3e.gif]]~~ <~~/center~~>	+	<math>e</math>
-	{{~~NAVCONTENT_STOP~~}}	+	raised to something” and therefore we differentiate using the chain rule:
		+
		+
		+	<math>\frac{d}{dx}e^{\left\{ \left. \sin x^{2} \right\} \right.}=e^{\left\{ \left. \sin x^{2} \right\} \right.}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }</math>
		+
		+
		+	Then, we differentiate “sine of something”:
		+
		+
		+	<math>\begin{align}
		+	& e^{\sin x^{2}}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }=e^{\sin x^{2}}\centerdot \cos \left\{ \left. x^{2} \right\} \right.\centerdot \left( \left\{ \left. x^{2} \right\} \right. \right)^{\prime } \\
		+	& =e^{\sin x^{2}}\centerdot \cos x^{2}\centerdot 2x \\
		+	\end{align}</math>

Tekbot: Lösning 1.2:3e moved to Solution 1.2:3e: Robot: moved page

2008-09-16T14:27:21Z

Lösning 1.2:3e moved to Solution 1.2:3e: Robot: moved page

← Nächstältere Version

Version vom 14:27, 16. Sep. 2008

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

2008-08-21T07:53:24Z

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

← Nächstältere Version		Version vom 07:53, 21. Aug. 2008
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-	<center> [[~~Bild~~:1_2_3e.gif]] </center>	+	<center> [[Image:1_2_3e.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_3e.gif {{NAVCONTENT_STOP}}

2008-04-04T08:46:51Z

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