Lösung 1.2:3f - Versionsgeschichte

Sekretariat1 um 11:00, 20. Aug. 2009

2009-08-20T11:00:35Z

← Nächstältere Version		Version vom 11:00, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Es gibt keine Regel um die Funktion direkt abzuleiten. Stattdessen verwenden wir die Regel	+	Es gibt keine Regel, um die Funktion direkt abzuleiten. Stattdessen verwenden wir die Regel

-	{{Abgesetzte Formel\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,.</math>}}

-	~~Dies~~ ergibt	+	Das ergibt

	{{Abgesetzte Formel\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}		{{Abgesetzte Formel\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}

-	Jetzt leiten ~~wor~~ die Funktion mit der Kettenregel ab	+	Jetzt leiten wir die Funktion mit der Kettenregel ab

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}</math>}}

-	und verwenden die Faktorregel,	+	und verwenden die Faktorregel

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 21:		Zeile 21:
	\end{align}</math>}}		\end{align}</math>}}

-	Wo wir (*) rückwärts verwendet haben.	+	wobei wir (*) rückwärts verwendet haben.

Martin um 12:30, 19. Apr. 2009

2009-04-19T12:30:51Z

← Nächstältere Version		Version vom 12:30, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We have no differentiation rule for a function raised to another function, but instead we use the formula~~	+	Es gibt keine Regel um die Funktion direkt abzuleiten. Stattdessen verwenden wir die Regel

	{{Abgesetzte Formel\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}		{{Abgesetzte Formel\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}

-	~~which, in our case, gives~~	+	Dies ergibt

	{{Abgesetzte Formel\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}		{{Abgesetzte Formel\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}

-	~~Now, we obtain the derivative by first using the chain rule~~	+	Jetzt leiten wor die Funktion mit der Kettenregel ab

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}</math>}}

-	~~and then the product rule~~	+	und verwenden die Faktorregel,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 21:		Zeile 21:
	\end{align}</math>}}		\end{align}</math>}}

-	~~where we have used~~ (*) ~~in reverse~~.	+	Wo wir (*) rückwärts verwendet haben.

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

2009-03-11T10:07:15Z

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

← Nächstältere Version

Version vom 10:07, 11. Mär. 2009

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

2009-03-10T12:54:44Z

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

← Nächstältere Version		Version vom 12:54, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We have no differentiation rule for a function raised to another function, but instead we use the formula		We have no differentiation rule for a function raised to another function, but instead we use the formula

-	{{~~Displayed math~~\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}

	which, in our case, gives		which, in our case, gives

-	{{~~Displayed math~~\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}	+	{{Abgesetzte Formel\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}

	Now, we obtain the derivative by first using the chain rule		Now, we obtain the derivative by first using the chain rule

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

	and then the product rule		and then the product rule

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{}		\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{}
	&= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt]		&= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt]

Tek um 13:24, 15. Okt. 2008

2008-10-15T13:24:50Z

← Nächstältere Version		Version vom 13:24, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	We have no differentiation rule for a function raised to another function, but instead we ~~rewrite~~	+	We have no differentiation rule for a function raised to another function, but instead we use the formula

-		+	{{Displayed math\|\|<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}
-	<math>a^{b}=e^{\ln a^{b}}=e^{b\ln a}</math>,	+

	which, in our case, gives		which, in our case, gives

-		+	{{Displayed math\|\|<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>\|(*)}}
-	<math>x^{\tan x}=e^{\tan x\~~centerdot~~ \ln x}</math>	+
-	(*)	+

	Now, we obtain the derivative by first using the chain rule		Now, we obtain the derivative by first using the chain rule

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

-	<math>\frac{d}{dx}e^{\left\{ \left. \tan x\centerdot \ln x \right\} \right.}=e^{\left\{ \left. \tan x\centerdot \ln x \right\} \right.}\centerdot \left( \left\{ \left. \tan x\centerdot \ln x \right\} \right. \right)^{\prime }</math>	+	and then the product rule
-		+
-		+
-	and then the product rule:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& =e^{\tan x\~~centerdot~~ \ln x}\~~left~~( ~~\left~~( \tan x ~~\right~~)^{\~~prime }\centerdot~~ \ln x+\tan x\~~centerdot \left~~( \ln x ~~\right~~)^{\~~prime } \right~~) \\	+	\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{}
-	& =e^{\tan x\~~centerdot~~ \ln x}\~~left~~( \frac{1}{\cos ^{2}x}\~~centerdot~~ \ln x+\tan x\~~centerdot~~ \frac{1}{x} \~~right~~) \\	+	&= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt]
-	& =e^{\tan x\~~centerdot~~ \ln x}\~~left~~( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \~~right~~) \\	+	&= e^{\tan x\cdot\ln x}\Bigl(\frac{1}{\cos^2\!x}\cdot\ln x + \tan x\cdot\frac{1}{x} \Bigr)\\[5pt]
-	& =x^{\tan x}\~~left~~( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \~~right~~) \\	+	&= e^{\tan x\cdot\ln x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\\[5pt]
-	\end{align}</math>	+	&= x^{\tan x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\,,
		+	\end{align}</math>}}

	where we have used (*) in reverse.		where we have used (*) in reverse.

Ian um 14:55, 13. Okt. 2008

2008-10-13T14:55:57Z

← Nächstältere Version		Version vom 14:55, 13. Okt. 2008
Zeile 8:		Zeile 8:

	<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>		<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
-		+	(*)

	Now, we obtain the derivative by first using the chain rule		Now, we obtain the derivative by first using the chain rule
Zeile 25:		Zeile 25:
	& =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\		& =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
	\end{align}</math>		\end{align}</math>
		+
		+	where we have used (*) in reverse.

Ian um 13:41, 12. Okt. 2008

2008-10-12T13:41:31Z

← Nächstältere Version		Version vom 13:41, 12. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We have no differentiation rule for a function raised to another function, but instead we rewrite
-	<~~center~~> ~~[[Image:1_2_3f~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>a^{b}=e^{\ln a^{b}}=e^{b\ln a}</math>,
		+
		+	which, in our case, gives
		+
		+
		+	<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
		+
		+
		+	Now, we obtain the derivative by first using the chain rule
		+
		+
		+	<math>\frac{d}{dx}e^{\left\{ \left. \tan x\centerdot \ln x \right\} \right.}=e^{\left\{ \left. \tan x\centerdot \ln x \right\} \right.}\centerdot \left( \left\{ \left. \tan x\centerdot \ln x \right\} \right. \right)^{\prime }</math>
		+
		+
		+	and then the product rule:
		+
		+
		+	<math>\begin{align}
		+	& =e^{\tan x\centerdot \ln x}\left( \left( \tan x \right)^{\prime }\centerdot \ln x+\tan x\centerdot \left( \ln x \right)^{\prime } \right) \\
		+	& =e^{\tan x\centerdot \ln x}\left( \frac{1}{\cos ^{2}x}\centerdot \ln x+\tan x\centerdot \frac{1}{x} \right) \\
		+	& =e^{\tan x\centerdot \ln x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
		+	& =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
		+	\end{align}</math>

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

2008-09-16T14:27:41Z

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

← Nächstältere Version

Version vom 14:27, 16. Sep. 2008

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

2008-08-21T07:53:34Z

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

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2008-04-04T08:47:26Z

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