Lösung 1.2:2e - Versionsgeschichte

Sekretariat1 um 10:12, 20. Aug. 2009

2009-08-20T10:12:18Z

← Nächstältere Version		Version vom 10:12, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Theoretisch ist es möglich den Ausdruck zu erweitern, und Term für Term abzuleiten. ~~Dies~~ ist aber mühsam, und wir verwenden daher stattdessen die Ableitungsregeln.	+	Theoretisch ist es möglich, den Ausdruck zu erweitern und Term für Term abzuleiten. Das ist aber mühsam und wir verwenden daher stattdessen die Ableitungsregeln.

	Durch die Faktorregel erhalten wir		Durch die Faktorregel erhalten wir
Zeile 13:		Zeile 13:
	{{Abgesetzte Formel\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}

-	Durch die ~~kettenregel~~ erhalten wir	+	Durch die Kettenregel erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 35:		Zeile 35:
	\end{align}</math>}}		\end{align}</math>}}

-	Wir hohlen schließlich den Faktor <math>(2x+1)^3</math> heraus,	+	Wir hohlen schließlich den Faktor <math>(2x+1)^3</math> heraus

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

Martin um 21:42, 18. Apr. 2009

2009-04-18T21:42:42Z

← Nächstältere Version		Version vom 21:42, 18. Apr. 2009
Zeile 1:		Zeile 1:
-	~~One way to differentiate the expression could be to expand <math>(2x+1)^4</math> multiply by <math>x</math> and differentiate term by term~~, ~~but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules~~.	+	Theoretisch ist es möglich den Ausdruck zu erweitern, und Term für Term abzuleiten. Dies ist aber mühsam, und wir verwenden daher stattdessen die Ableitungsregeln.

-	~~To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that~~	+	Durch die Faktorregel erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 9:		Zeile 9:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We can differentiate the expression~~ <math>(2x+1)^4</math> ~~by viewing it as "something raised to the 4"~~,	+	Wir berechnen die Ableitung von <math>(2x+1)^4</math> mit der Kettenregel, indem wir folgenden Ausdruck betrachten

	{{Abgesetzte Formel\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}

-	~~The chain rule then gives~~	+	Durch die kettenregel erhalten wir

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

-	~~We carry out the last differentiation directly, and obtain~~	+	Die letzte Ableitung ist einfach

	{{Abgesetzte Formel\|\|<math>(2x+1)' = 2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(2x+1)' = 2\,\textrm{.}</math>}}

-	~~If we go through the whole calculation from the beginning~~, ~~it is~~	+	Machen wir alle Schritte von Anfang an, erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 35:		Zeile 35:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Both terms contain a common factor~~ <math>(2x+1)^3</math> ~~which we can take out to get an answer in factorized form~~,	+	Wir hohlen schließlich den Faktor <math>(2x+1)^3</math> heraus,

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

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

2009-03-11T10:04:55Z

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

← Nächstältere Version

Version vom 10:04, 11. Mär. 2009

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

2009-03-10T12:53:34Z

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

← Nächstältere Version		Version vom 12:53, 10. Mär. 2009
Zeile 3:		Zeile 3:
	To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that		To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]		\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
	&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]		&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
Zeile 11:		Zeile 11:
	We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",		We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",

-	{{~~Displayed math~~\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}

	The chain rule then gives		The chain rule then gives

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]		\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
	&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]		&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]
Zeile 23:		Zeile 23:
	We carry out the last differentiation directly, and obtain		We carry out the last differentiation directly, and obtain

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

	If we go through the whole calculation from the beginning, it is		If we go through the whole calculation from the beginning, it is

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]		\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
	&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]		&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]
Zeile 37:		Zeile 37:
	Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,		Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]		\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
	&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]		&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]
	&= (2x+1)^3(10x+1)\,\textrm{.}		&= (2x+1)^3(10x+1)\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 08:56, 15. Okt. 2008

2008-10-15T08:56:12Z

← Nächstältere Version		Version vom 08:56, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	One way to differentiate the expression could be to expand	+	One way to differentiate the expression could be to expand <math>(2x+1)^4</math> multiply by <math>x</math> and differentiate term by term, but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules.
-	<math>~~\left~~( 2x+1 ~~\right~~)^{4}</math>	+
-	multiply by	+
-	<math>x</math>	+
-	and differentiate term by term, but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules.	+

-	To begin with, we have a product of	+	To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that
-	<math>x</math>	+
-	and	+
-	<math>~~\left~~( 2x+1 ~~\right~~)^{4}</math>	+
-	so the product rule gives that	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
		+	&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
		+	&= 1\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\,\textrm{.}
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
-	~~& \frac{d}{dx}x\left~~( 2x+1 ~~\right~~)^{4~~}=\left( x \right)^{\prime }\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\~~	+
-	~~& =1\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\~~	+
-	~~\end{align}~~</math>	+

-		+	{{Displayed math\|\|<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
-		+
-	~~We can differentiate the expression~~	+
-	<math>\~~left~~( 2x+1 \~~right)~~^{4}~~</math>~~	+
-	~~by viewing it as "something raised to the~~	+
-	~~<math>~~\~~text{4}</math>"~~,	+
-		+
-		+
-	~~<math>\left~~\{ ~~\left~~. ~~{} \right\} \right.^{4~~}</math>.	+

	The chain rule then gives		The chain rule then gives

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
-	& \frac{d}{dx}\~~left~~\{ \~~left~~. {} \~~right~~\} ~~\right.~~^{4}=4\~~centerdot \left~~\{ \~~left.~~ {} \~~right~~\} ~~\right.~~^{3}\~~centerdot \left( \left~~\{ \~~left.~~ {} \~~right~~\} ~~\right. \right)~~^{\prime } \\	+	&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]
-	& \frac{d}{dx}\~~left~~( 2x+1 ~~\right~~)^{4}=4\~~centerdot \left~~( 2x+1 ~~\right~~)^{3}\~~centerdot \left~~( 2x+1 ~~\right~~)~~^{\prime }~~ \\	+	\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}
-		+

	We carry out the last differentiation directly, and obtain		We carry out the last differentiation directly, and obtain

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

	If we go through the whole calculation from the beginning, it is		If we go through the whole calculation from the beginning, it is

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
		+	&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]
		+	&= 1\cdot (2x+1)^4 + x\cdot 4(2x+1)^3\cdot (2x+1)'\\[5pt]
		+	&= (2x+1)^4 + x\cdot 4(2x+1)^3\cdot 2\\[5pt]
		+	&= (2x+1)^4 + 8x(2x+1)^3\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
-	~~& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( x \right)^{\prime }\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\~~	+
-	~~& =1\centerdot \left( 2x+1 \right)^{4}+x\centerdot 4\left( 2x+1 \right)^{3}\centerdot \left( 2x+1 \right)^{\prime } \\~~	+
-	~~& =\left( 2x+1 \right)^{4}+x\centerdot 4\left( 2x+1 \right)^{3}\centerdot 2 \\~~	+
-	~~& =\left( 2x+1 \right)^{4}+8x\left( 2x+1 \right)^{3} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Both terms contain a common factor	+
-	<math>~~\left~~( 2x+1 ~~\right~~)^{3}</math>	+
-	which we can take out to get an answer in factorized form:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{d}{dx}x~~\left~~( 2x+1 ~~\right~~)^{4}=~~\left~~( 2x+1 ~~\right~~)^{3}\~~left~~( ~~\left~~( 2x+1 ~~\right~~)+8x \~~right~~) \\	+	\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
-	& =~~\left~~( 2x+1 ~~\right~~)^{3~~}\left~~( 10x+1 ~~\right~~) \\	+	&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]
-	\end{align}</math>	+	&= (2x+1)^3(10x+1)\,\textrm{.}
		+	\end{align}</math>}}

Ian um 13:39, 11. Okt. 2008

2008-10-11T13:39:13Z

← Nächstältere Version		Version vom 13:39, 11. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	One way to differentiate the expression could be to expand
-	<~~center~~> ~~[[Image:1_2_2e-~~1(2).~~gif]]~~ </~~center~~>	+	<math>\left( 2x+1 \right)^{4}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	multiply by
-	{{~~NAVCONTENT_START~~}}	+	<math>x</math>
-	<~~center~~> ~~[[Image:1_2_2e-~~2(2)~~.gif]]~~ </~~center~~>	+	and differentiate term by term, but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules.
-	{{~~NAVCONTENT_STOP~~}}	+
		+	To begin with, we have a product of
		+	<math>x</math>
		+	and
		+	<math>\left( 2x+1 \right)^{4}</math>
		+	so the product rule gives that
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( x \right)^{\prime }\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\
		+	& =1\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\
		+	\end{align}</math>
		+
		+
		+
		+	We can differentiate the expression
		+	<math>\left( 2x+1 \right)^{4}</math>
		+	by viewing it as "something raised to the
		+	<math>\text{4}</math>",
		+
		+
		+	<math>\left\{ \left. {} \right\} \right.^{4}</math>.
		+
		+	The chain rule then gives
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\left\{ \left. {} \right\} \right.^{4}=4\centerdot \left\{ \left. {} \right\} \right.^{3}\centerdot \left( \left\{ \left. {} \right\} \right. \right)^{\prime } \\
		+	& \frac{d}{dx}\left( 2x+1 \right)^{4}=4\centerdot \left( 2x+1 \right)^{3}\centerdot \left( 2x+1 \right)^{\prime } \\
		+	\end{align}</math>
		+
		+
		+	We carry out the last differentiation directly, and obtain
		+
		+
		+	<math>\left( 2x+1 \right)^{\prime }=2</math>
		+
		+
		+	If we go through the whole calculation from the beginning, it is
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( x \right)^{\prime }\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\
		+	& =1\centerdot \left( 2x+1 \right)^{4}+x\centerdot 4\left( 2x+1 \right)^{3}\centerdot \left( 2x+1 \right)^{\prime } \\
		+	& =\left( 2x+1 \right)^{4}+x\centerdot 4\left( 2x+1 \right)^{3}\centerdot 2 \\
		+	& =\left( 2x+1 \right)^{4}+8x\left( 2x+1 \right)^{3} \\
		+	\end{align}</math>
		+
		+
		+	Both terms contain a common factor
		+	<math>\left( 2x+1 \right)^{3}</math>
		+	which we can take out to get an answer in factorized form:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( 2x+1 \right)^{3}\left( \left( 2x+1 \right)+8x \right) \\
		+	& =\left( 2x+1 \right)^{3}\left( 10x+1 \right) \\
		+	\end{align}</math>

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

2008-09-16T14:25:21Z

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

← Nächstältere Version

Version vom 14:25, 16. Sep. 2008

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

2008-08-21T07:52:24Z

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

← Nächstältere Version		Version vom 07:52, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_2e-1(2).gif]] </center>	+	<center> [[Image:1_2_2e-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_2e-2(2).gif]] </center>	+	<center> [[Image:1_2_2e-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_2e-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:1_2_2e-2(2).gif {{NAVCONTENT_STOP}}

2008-04-04T08:38:16Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2e-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_2e-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:1_2_2e-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:1_2_2e-2(2).gif]] </center>
{{NAVCONTENT_STOP}}