Lösung 1.2:2f - Versionsgeschichte

Sekretariat1 um 10:14, 20. Aug. 2009

2009-08-20T10:14:31Z

← Nächstältere Version		Version vom 10:14, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Die Funktion ist mehrmals verkettet,	+	Die Funktion ist mehrmals verkettet

	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}		{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}

-	Die äußerste Ableitung erhalten wir indem wir den Ausdruck	+	Die äußerste Ableitung erhalten wir, indem wir den Ausdruck

-	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,</math>}}

-	mit der Kettenregel ableiten,	+	mit der Kettenregel ableiten.

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

-	Im nächsten Schritt haben wir die verkettete Funktion "die Wurzel von ...",	+	Im nächsten Schritt haben wir die verkettete Funktion "die Wurzel von ..."

	{{Abgesetzte Formel\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}		{{Abgesetzte Formel\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}

Martin um 21:45, 18. Apr. 2009

2009-04-18T21:45:39Z

← Nächstältere Version		Version vom 21:45, 18. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The entire expression is made up of several levels~~,	+	Die Funktion ist mehrmals verkettet,

	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}		{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}

-	~~and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",~~	+	Die äußerste Ableitung erhalten wir indem wir den Ausdruck

	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}		{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}

-	~~and differentiate this using the chain rule~~,	+	mit der Kettenregel ableiten,

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}</math>}}

-	~~In the next differentiation, we have~~ "~~the square root of something~~",	+	Im nächsten Schritt haben wir die verkettete Funktion "die Wurzel von ...",

	{{Abgesetzte Formel\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}		{{Abgesetzte Formel\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}

-	~~where we have used the differentiation rule,~~	+	wo wir

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}

-	~~for the outer derivative~~.	+	verwendet haben.

-	~~The whole differentiation in one go becomes~~	+	Die Ableitung der ganzen Funktion ist also

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

Tekbot: Solution 1.2:2f moved to Lösung 1.2:2f: Robot: moved page

2009-03-11T10:05:15Z

Solution 1.2:2f moved to Lösung 1.2:2f: Robot: moved page

← Nächstältere Version

Version vom 10:05, 11. Mär. 2009

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

2009-03-10T12:53:44Z

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

Tek um 10:49, 15. Okt. 2008

2008-10-15T10:49:35Z

← Nächstältere Version		Version vom 10:49, 15. Okt. 2008
Zeile 1:		Zeile 1:
	The entire expression is made up of several levels,		The entire expression is made up of several levels,

		+	{{Displayed math\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}

-	~~<math>\cos \left\{ \left~~. ~~\sqrt{\left\{ \left. 1-x \right\} \right.} \right\} \right.</math>~~	+	and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",

		+	{{Displayed math\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}

-	and ~~when we~~ differentiate ~~we go from~~ the ~~outside inwards. In the first stage, we consider the expression as "cos of something"~~,	+	and differentiate this using the chain rule,

		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}</math>}}

-	~~<math>\cos \left\{ \left. {} \right\} \right.</math>~~	+	In the next differentiation, we have "the square root of something",
-		+
-		+
-	~~and differentiate this using the chain rule:~~	+
-		+
-		+
-	~~<math>\frac{d}{dx}\cos \left\{ \left. \sqrt{1-x} \right\} \right.=-\sin \left\{ \left. \sqrt{1-x} \right\} \right.\centerdot \left( \left\{ \left. \sqrt{1-x} \right\} \right. \right)^{\prime }</math>~~	+
-		+
-		+
-	In the next differentiation, we have "the root of something",	+
-		+
-		+
-	~~<math>\left( \sqrt{\left\{ \left. 1-x \right\} \right.} \right)^{\prime }=\frac{1}{2\sqrt{1-x}}\centerdot \left( 1-x \right)^{\prime }</math>~~	+

		+	{{Displayed math\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}

	where we have used the differentiation rule,		where we have used the differentiation rule,

-		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
-	<math>\frac{d}{dx}\~~left~~( \sqrt{x} \~~right~~)=\frac{1}{2\sqrt{x}}</math>	+
-		+

	for the outer derivative.		for the outer derivative.

-	The whole differentiation in one go becomes:	+	The whole differentiation in one go becomes
-		+

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

Ian um 13:53, 11. Okt. 2008

2008-10-11T13:53:03Z

← Nächstältere Version		Version vom 13:53, 11. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The entire expression is made up of several levels,
-	<~~center~~> ~~[[Image~~:~~1_2_2f~~-1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\cos \left\{ \left. \sqrt{\left\{ \left. 1-x \right\} \right.} \right\} \right.</math>
-	<~~center~~> ~~[[Image~~:~~1_2_2f~~-2(2)~~.gif]] </center>~~	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cos of something",
		+
		+
		+	<math>\cos \left\{ \left. {} \right\} \right.</math>
		+
		+
		+	and differentiate this using the chain rule:
		+
		+
		+	<math>\frac{d}{dx}\cos \left\{ \left. \sqrt{1-x} \right\} \right.=-\sin \left\{ \left. \sqrt{1-x} \right\} \right.\centerdot \left( \left\{ \left. \sqrt{1-x} \right\} \right. \right)^{\prime }</math>
		+
		+
		+	In the next differentiation, we have "the root of something",
		+
		+
		+	<math>\left( \sqrt{\left\{ \left. 1-x \right\} \right.} \right)^{\prime }=\frac{1}{2\sqrt{1-x}}\centerdot \left( 1-x \right)^{\prime }</math>
		+
		+
		+	where we have used the differentiation rule,
		+
		+
		+	<math>\frac{d}{dx}\left( \sqrt{x} \right)=\frac{1}{2\sqrt{x}}</math>
		+
		+
		+	for the outer derivative.
		+
		+	The whole differentiation in one go becomes:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\cos \sqrt{1-x}=-\sin \sqrt{1-x}\centerdot \frac{d}{dx}\sqrt{1-x} \\
		+	& =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \frac{d}{dx}\left( 1-x \right) \\
		+	& =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \left( -1 \right) \\
		+	& =\frac{\sin \sqrt{1-x}}{2\sqrt{1-x}} \\
		+	\end{align}</math>

Tekbot: Lösning 1.2:2f moved to Solution 1.2:2f: Robot: moved page

2008-09-16T14:25:41Z

Lösning 1.2:2f moved to Solution 1.2:2f: Robot: moved page

← Nächstältere Version

Version vom 14:25, 16. Sep. 2008

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

2008-08-21T07:52:34Z

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

← Nächstältere Version		Version vom 07:52, 21. Aug. 2008
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-	<center> [[~~Bild~~:1_2_2f-1(2).gif]] </center>	+	<center> [[Image:1_2_2f-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_2f-2(2).gif]] </center>	+	<center> [[Image:1_2_2f-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_2f-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:1_2_2f-2(2).gif {{NAVCONTENT_STOP}}

2008-04-04T08:39:08Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2f-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_2f-2(2).gif </center> {{NAVCONTENT_STOP}}

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← Nächstältere Version		Version vom 12:53, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The entire expression is made up of several levels,		The entire expression is made up of several levels,

-	{{~~Displayed math~~\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}

	and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",		and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",

-	{{~~Displayed math~~\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}

	and differentiate this using the chain rule,		and differentiate this using the chain rule,

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

	In the next differentiation, we have "the square root of something",		In the next differentiation, we have "the square root of something",

-	{{~~Displayed math~~\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}

	where we have used the differentiation rule,		where we have used the differentiation rule,

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

	for the outer derivative.		for the outer derivative.
Zeile 23:		Zeile 23:
	The whole differentiation in one go becomes		The whole differentiation in one go becomes

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\cos\sqrt{1-x}		\frac{d}{dx}\cos\sqrt{1-x}
	&= -\sin\sqrt{1-x}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt]		&= -\sin\sqrt{1-x}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt]