Lösung 1.2:3b - Versionsgeschichte

Sekretariat1 um 10:34, 20. Aug. 2009

2009-08-20T10:34:33Z

← Nächstältere Version		Version vom 10:34, 20. Aug. 2009
Zeile 7:		Zeile 7:
	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}

-	Die innere Ableitung berechnen wir mit der Quotientenregel,	+	Die innere Ableitung berechnen wir mit der Quotientenregel

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 18:		Zeile 18:
	\end{align}</math>}}		\end{align}</math>}}

-	Wo wir die Vereinfachungen	+	wo wir die Vereinfachungen

	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}		{{Abgesetzte Formel\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}
Zeile 24:		Zeile 24:
	= (x-1)^{1/2-2}		= (x-1)^{1/2-2}
	= (x-1)^{-3/2}		= (x-1)^{-3/2}
-	= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}	+	= \frac{1}{(x-1)^{3/2}}\,\textrm{}</math>}}

	verwendet haben.		verwendet haben.

Martin um 11:33, 19. Apr. 2009

2009-04-19T11:33:22Z

← Nächstältere Version		Version vom 11:33, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The outer function in the expression is "the square root of something",~~	+	Die äußere Funktion ist

	{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}

-	~~and differentiating with the chain rule gives~~	+	und mit der Kettenregel erhalten wir die Ableitung

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}

-	~~We establish the inner derivative by using the quotient rule~~,	+	Die innere Ableitung berechnen wir mit der Quotientenregel,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 18:		Zeile 18:
	\end{align}</math>}}		\end{align}</math>}}

-	~~where we have used the simplification~~	+	Wo wir die Vereinfachungen

	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}		{{Abgesetzte Formel\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}
Zeile 25:		Zeile 25:
	= (x-1)^{-3/2}		= (x-1)^{-3/2}
	= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}		= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}
		+
		+	verwendet haben.

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

2009-03-11T10:05:55Z

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

← Nächstältere Version

Version vom 10:05, 11. Mär. 2009

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

2009-03-10T12:54:04Z

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

← Nächstältere Version		Version vom 12:54, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The outer function in the expression is "the square root of something",		The outer function in the expression is "the square root of something",

-	{{~~Displayed math~~\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}

	and differentiating with the chain rule gives		and differentiating with the chain rule gives

-	{{~~Displayed math~~\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}

	We establish the inner derivative by using the quotient rule,		We establish the inner derivative by using the quotient rule,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}		\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
	&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt]		&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt]
Zeile 20:		Zeile 20:
	where we have used the simplification		where we have used the simplification

-	{{~~Displayed math~~\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}	+	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}
	= \frac{(x-1)^{1/2}}{(x-1)^2}		= \frac{(x-1)^{1/2}}{(x-1)^2}
	= (x-1)^{1/2-2}		= (x-1)^{1/2-2}
	= (x-1)^{-3/2}		= (x-1)^{-3/2}
	= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}		= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}

Tek um 11:53, 15. Okt. 2008

2008-10-15T11:53:54Z

← Nächstältere Version		Version vom 11:53, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	The outer function in the expression is "the root of something",	+	The outer function in the expression is "the square root of something",
-		+
-		+
-	~~<math>\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}</math>~~	+

		+	{{Displayed math\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}

	and differentiating with the chain rule gives		and differentiating with the chain rule gives

-		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}
-	<math>\frac{d}{dx}\sqrt{\~~left\{ \left~~. \frac{x+1}{x-1} ~~\right\~~} ~~\right.~~}=\frac{1}{2\sqrt{\~~left\~~{ \~~left. \frac~~{x+1}{x-1} ~~\right\~~} ~~\right.~~}}\~~centerdot~~ \~~left~~( \frac{x+1}{x-1} \~~right~~)^{\~~prime~~ }</math>	+
-		+

	We establish the inner derivative by using the quotient rule,		We establish the inner derivative by using the quotient rule,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
-	& \frac{d}{dx}\sqrt{\frac{x+1}{x-1}}=\frac{1}{2\sqrt{\~~frac~~{x+1}{x-1}}}\~~centerdot~~ \frac{~~\left~~( x+1 ~~\right~~)~~^{\prime }\centerdot~~ \~~left~~( x-1 ~~\right~~)-~~\left~~( x+1 ~~\right~~)\~~centerdot \left~~( x-1 ~~\right~~)~~^{\prime }~~}{~~\left~~( x-1 ~~\right~~)^{2}} \\	+	&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt]
-	& =\frac{1}{2\sqrt{\~~frac~~{x+1}{x-1}}}\~~centerdot~~ \frac{1\~~centerdot \left~~( x-1 ~~\right~~)-~~\left~~( x+1 ~~\right~~)\~~centerdot~~ 1}{~~\left~~( x-1 ~~\right~~)^{2}} \\	+	&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{1\cdot (x-1) - (x+1)\cdot 1}{(x-1)^2}\\[5pt]
-	& =\frac{1}{2\sqrt{\~~frac~~{x+1}{x-1}}}\~~centerdot~~ \frac{-2}{~~\left~~( x-1 ~~\right~~)^{2}} \\	+	&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt]
-	& =-\sqrt{\frac{x-1}{x+1}~~\centerdot~~ }\frac{1}{~~\left~~( x-1 ~~\right~~)^{2}} \\	+	&= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt]
-	& =-\frac{1}{~~\left~~( x-1 ~~\right~~)^{{3}/{2~~}\;~~}\sqrt{x+1}} \\	+	&= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,,
-	\end{align}</math>	+	\end{align}</math>}}
-		+

	where we have used the simplification		where we have used the simplification

-		+	{{Displayed math\|\|<math>\frac{\sqrt{x-1}}{(x-1)^2}
-	<math>\frac{\sqrt{x-1}}{~~\left~~( x-1 ~~\right~~)^{2}}=\frac{~~\left~~( x-1 ~~\right~~)^{{1}/{2~~}\;~~}}{~~\left~~( x-1 ~~\right~~)^{2}}=~~\left~~( x-1 ~~\right~~)^~~{\frac~~{1}{2}-2}=~~\left~~( x-1 ~~\right~~)^{-~~\frac{~~3}{2}}=\frac{1}{~~\left~~( x-1 ~~\right~~)^~~{\frac~~{3}{2}}}</math>	+	= \frac{(x-1)^{1/2}}{(x-1)^2}
		+	= (x-1)^{1/2-2}
		+	= (x-1)^{-3/2}
		+	= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}

Ian um 12:45, 12. Okt. 2008

2008-10-12T12:45:22Z

← Nächstältere Version		Version vom 12:45, 12. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The outer function in the expression is "the root of something",
-	<~~center~~> ~~[[Image:1_2_3b~~-1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}</math>
-	~~<center> [[Image:1_2_3b~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	and differentiating with the chain rule gives
		+
		+
		+	<math>\frac{d}{dx}\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}}\centerdot \left( \frac{x+1}{x-1} \right)^{\prime }</math>
		+
		+
		+	We establish the inner derivative by using the quotient rule,
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\sqrt{\frac{x+1}{x-1}}=\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{\left( x+1 \right)^{\prime }\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot \left( x-1 \right)^{\prime }}{\left( x-1 \right)^{2}} \\
		+	& =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{1\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot 1}{\left( x-1 \right)^{2}} \\
		+	& =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{-2}{\left( x-1 \right)^{2}} \\
		+	& =-\sqrt{\frac{x-1}{x+1}\centerdot }\frac{1}{\left( x-1 \right)^{2}} \\
		+	& =-\frac{1}{\left( x-1 \right)^{{3}/{2}\;}\sqrt{x+1}} \\
		+	\end{align}</math>
		+
		+
		+	where we have used the simplification
		+
		+
		+	<math>\frac{\sqrt{x-1}}{\left( x-1 \right)^{2}}=\frac{\left( x-1 \right)^{{1}/{2}\;}}{\left( x-1 \right)^{2}}=\left( x-1 \right)^{\frac{1}{2}-2}=\left( x-1 \right)^{-\frac{3}{2}}=\frac{1}{\left( x-1 \right)^{\frac{3}{2}}}</math>

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

2008-09-16T14:26:21Z

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

← Nächstältere Version

Version vom 14:26, 16. Sep. 2008

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

2008-08-21T07:52:54Z

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

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2008-04-04T08:41:12Z

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