Lösung 1.2:2b - Versionsgeschichte

Sekretariat1 um 07:42, 20. Aug. 2009

2009-08-20T07:42:49Z

← Nächstältere Version		Version vom 07:42, 20. Aug. 2009
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}</math>}}

-	Die innere Funktion ist ein Polynom, und wir erhalten direkt die innere Ableitung,	+	Die innere Funktion ist ein Polynom und wir erhalten direkt die innere Ableitung

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}

Martin um 20:54, 18. Apr. 2009

2009-04-18T20:54:16Z

← Nächstältere Version		Version vom 20:54, 18. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The whole expression consists of two parts: the outer part, "''e'' raised to something"~~,	+	Die Funktion besteht aus der äußeren Exponentialfunktion,

	{{Abgesetzte Formel\|\|<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}		{{Abgesetzte Formel\|\|<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}

-	~~where "something" is the inner part~~ <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.	+	und der inneren Funktion <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.

-	~~The derivative is calculated according to the chain rule by differentiating~~ <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> ~~with respect to~~ <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> ~~and then multiplying by the inner derivative~~ <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math>, ~~i.e.~~	+	Wir erhalten die Ableitung der Funktion mit der Kettenregel, indem wir die Ableitung von <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> in Bezug auf <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> mit der inneren Ableitung <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math> multiplizieren, also

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

-	~~The inner part is an ordinary polynomial which we differentiate directly~~,	+	Die innere Funktion ist ein Polynom, und wir erhalten direkt die innere Ableitung,

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}

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

2009-03-11T10:03:55Z

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

← Nächstältere Version

Version vom 10:03, 11. Mär. 2009

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

2009-03-10T12:53:04Z

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

← Nächstältere Version		Version vom 12:53, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The whole expression consists of two parts: the outer part, "''e'' raised to something",		The whole expression consists of two parts: the outer part, "''e'' raised to something",

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

	where "something" is the inner part <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.		where "something" is the inner part <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.
Zeile 7:		Zeile 7:
	The derivative is calculated according to the chain rule by differentiating <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> with respect to <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> and then multiplying by the inner derivative <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math>, i.e.		The derivative is calculated according to the chain rule by differentiating <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> with respect to <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> and then multiplying by the inner derivative <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math>, i.e.

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

	The inner part is an ordinary polynomial which we differentiate directly,		The inner part is an ordinary polynomial which we differentiate directly,

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

Tek um 08:02, 15. Okt. 2008

2008-10-15T08:02:18Z

← Nächstältere Version		Version vom 08:02, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	The whole expression consists of two parts: the outer part, "	+	The whole expression consists of two parts: the outer part, "''e'' raised to something",
-	~~<math>~~e~~</math>~~	+
-	raised to something",	+

		+	{{Displayed math\|\|<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}

-	<math>~~e^{\left\{~~ \~~left~~. {} \~~right~~\} \~~right.~~}</math>	+	where "something" is the inner part <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.

		+	The derivative is calculated according to the chain rule by differentiating <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> with respect to <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> and then multiplying by the inner derivative <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math>, i.e.

-	~~where "something" is the inner part~~	+	{{Displayed math\|\|<math>\frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}</math>}}
-	<math>\~~left\~~{ ~~\left.~~ {} \~~right~~\} \~~right~~.=x^{2}+x~~</math>. The derivative is calculated according to the chain rule by differentiating~~	+
-	~~<math>~~e^{\~~left~~\~~{ \left~~. {} \~~right~~\} ~~\right.~~}~~</math>~~	+
-	~~with respect to~~	+
-	~~<math>~~\~~left~~\{ \~~left~~. {} \~~right~~\} \~~right.</math>~~	+
-	~~and then multiplying by the inner derivative~~	+
-	~~<math>~~\~~left( \left~~\{ ~~\left~~. ~~{} \right\} \right. \right)^{\prime~~ }</math>~~, i.e.~~	+

		+	The inner part is an ordinary polynomial which we differentiate directly,

-	<math>\frac{d}{dx}e^{~~\left\{ \left.~~ x^{2}+x ~~\right\} \right.~~}=e^{~~\left\{ \left.~~ x^{2}+x ~~\right\~~} \~~right.}\centerdot \left~~( \~~left~~\{ ~~\left~~. ~~x^{2}+x \right\} \right. \right)^{\prime~~ }</math>	+	{{Displayed math\|\|<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}
-		+
-		+
-	~~The inner part is an ordinary polynomial which we differentiate directly:~~	+
-		+
-		+
-	~~<math>\frac{d~~}~~{dx~~}~~e^{\left\{ \left. x^{2}+x \right\} \right.}=e^{\left\{ \left. x^{2}+x \right\} \right.}\centerdot \left( 2x+1 \right)</math>~~	+

Ian um 12:56, 11. Okt. 2008

2008-10-11T12:56:03Z

← Nächstältere Version		Version vom 12:56, 11. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The whole expression consists of two parts: the outer part, "
-	<~~center~~> ~~[[Image:1_2_2b~~.~~gif]]~~ </~~center~~>	+	<math>e</math>
-	{{~~NAVCONTENT_STOP~~}}	+	raised to something",
		+
		+
		+	<math>e^{\left\{ \left. {} \right\} \right.}</math>
		+
		+
		+	where "something" is the inner part
		+	<math>\left\{ \left. {} \right\} \right.=x^{2}+x</math>. The derivative is calculated according to the chain rule by differentiating
		+	<math>e^{\left\{ \left. {} \right\} \right.}</math>
		+	with respect to
		+	<math>\left\{ \left. {} \right\} \right.</math>
		+	and then multiplying by the inner derivative
		+	<math>\left( \left\{ \left. {} \right\} \right. \right)^{\prime }</math>, i.e.
		+
		+
		+	<math>\frac{d}{dx}e^{\left\{ \left. x^{2}+x \right\} \right.}=e^{\left\{ \left. x^{2}+x \right\} \right.}\centerdot \left( \left\{ \left. x^{2}+x \right\} \right. \right)^{\prime }</math>
		+
		+
		+	The inner part is an ordinary polynomial which we differentiate directly:
		+
		+
		+	<math>\frac{d}{dx}e^{\left\{ \left. x^{2}+x \right\} \right.}=e^{\left\{ \left. x^{2}+x \right\} \right.}\centerdot \left( 2x+1 \right)</math>

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

2008-09-16T14:24:21Z

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

← Nächstältere Version

Version vom 14:24, 16. Sep. 2008

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

2008-08-21T07:51:54Z

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

← Nächstältere Version		Version vom 07:51, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_2b.gif]] </center>	+	<center> [[Image:1_2_2b.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_2b.gif {{NAVCONTENT_STOP}}

2008-04-04T08:36:07Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2b.gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:1_2_2b.gif]] </center>
{{NAVCONTENT_STOP}}