Lösung 1.2:1c - Versionsgeschichte

Sekretariat1 um 14:23, 19. Aug. 2009

Sekretariat1 — Wed, 19 Aug 2009 14:23:34 GMT

← Nächstältere Version		Version vom 14:23, 19. Aug. 2009
Zeile 1:		Zeile 1:
-	Der ~~ausdruck~~ ist ein Bruch mit den Zähler <math>x^2+1</math> und ~~den nenner~~ <math>x+1</math>, ~~und~~ daher verwenden wir die Quotientenregel um die Funktion abzuleiten,	+	Der Ausdruck ist ein Bruch mit den Zähler <math>x^2+1</math> und dem Nenner <math>x+1</math>, daher verwenden wir die Quotientenregel, um die Funktion abzuleiten.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 6:		Zeile 6:
	&= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt]		&= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt]
	&= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt]		&= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt]
-	&= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{.}	+	&= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

-	Hinweis: Wir können den Zähler durch quadratische Ergänzung umschreiben,	+	Hinweis: Wir können den Zähler durch quadratische Ergänzung umschreiben.

	{{Abgesetzte Formel\|\|<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}		{{Abgesetzte Formel\|\|<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}

-	~~und~~ erhalten ~~dadurch~~	+	Dadurch erhalten wir

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

Martin um 19:10, 18. Apr. 2009

Martin — Sat, 18 Apr 2009 19:10:45 GMT

← Nächstältere Version		Version vom 19:10, 18. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The expression is a quotient of two polynomials,~~ <math>x^2+1</math> ~~and~~ <math>x+1</math>, ~~and we therefore use the quotient rule for differentiation~~,	+	Der ausdruck ist ein Bruch mit den Zähler <math>x^2+1</math> und den nenner <math>x+1</math>, und daher verwenden wir die Quotientenregel um die Funktion abzuleiten,

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

-		+	Hinweis: Wir können den Zähler durch quadratische Ergänzung umschreiben,
-	~~Note~~: ~~It is possible to rewrite the numerator by completing the square~~,	+

	{{Abgesetzte Formel\|\|<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}		{{Abgesetzte Formel\|\|<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}

-	~~and then the answer can be written as~~	+	und erhalten dadurch

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

Tekbot: Solution 1.2:1c moved to Lösung 1.2:1c: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:02:15 GMT

Solution 1.2:1c moved to Lösung 1.2:1c: Robot: moved page

← Nächstältere Version

Version vom 10:02, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 12:52:14 GMT

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

← Nächstältere Version		Version vom 12:52, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The expression is a quotient of two polynomials, <math>x^2+1</math> and <math>x+1</math>, and we therefore use the quotient rule for differentiation,		The expression is a quotient of two polynomials, <math>x^2+1</math> and <math>x+1</math>, and we therefore use the quotient rule for differentiation,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\Bigl(\frac{x^2+1}{x+1}\Bigr)'		\Bigl(\frac{x^2+1}{x+1}\Bigr)'
	&= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt]		&= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt]
Zeile 12:		Zeile 12:
	Note: It is possible to rewrite the numerator by completing the square,		Note: It is possible to rewrite the numerator by completing the square,

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

	and then the answer can be written as		and then the answer can be written as

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

Tek um 13:51, 14. Okt. 2008

Tek — Tue, 14 Oct 2008 13:51:29 GMT

← Nächstältere Version		Version vom 13:51, 14. Okt. 2008
Zeile 1:		Zeile 1:
-	The expression is a quotient of two polynomials,	+	The expression is a quotient of two polynomials, <math>x^2+1</math> and <math>x+1</math>, and we therefore use the quotient rule for differentiation,
-	<math>x^{2}+1</math>	+
-	and	+
-	<math>x+1</math>, and we therefore use the quotient rule for differentiation:	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\Bigl(\frac{x^2+1}{x+1}\Bigr)'
		+	&= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt]
		+	&= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt]
		+	&= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt]
		+	&= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{.}
		+	\end{align}</math>}}

-	<math>\begin{align}
-	& \left( \frac{x^{2}+1}{x+1} \right)^{\prime }=\frac{\left( x^{2}+1 \right)^{\prime }\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot \left( x+1 \right)^{\prime }}{\left( x+1 \right)^{2}} \\
-	& \\
-	& =\frac{2x\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot 1}{\left( x+1 \right)^{2}} \\
-	& \\
-	& =\frac{2x^{2}+2x-x^{2}-1}{\left( x+1 \right)^{2}}=\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}} \\
-	\end{align}</math>

		+	Note: It is possible to rewrite the numerator by completing the square,

-	~~NOTE: it is possible to rewrite the numerator by completing the square,~~	+	{{Displayed math\|\|<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}
-		+
-		+
-	<math>x^{2}+2x-1=~~\left~~( x+1 ~~\right~~)^{2}-1^{2}-1=~~\left~~( x+1 ~~\right~~)^{2}-2</math>	+
-		+

	and then the answer can be written as		and then the answer can be written as

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

Tekbot: Lösning 1.2:1c moved to Solution 1.2:1c: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:22:41 GMT

Lösning 1.2:1c moved to Solution 1.2:1c: Robot: moved page

← Nächstältere Version

Version vom 14:22, 16. Sep. 2008

Ian um 15:06, 12. Sep. 2008

Ian — Fri, 12 Sep 2008 15:06:44 GMT

← Nächstältere Version		Version vom 15:06, 12. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The expression is a quotient of two polynomials,
-	<~~center~~> ~~[[Image~~:~~1_2_1c.gif]]~~ </~~center~~>	+	<math>x^{2}+1</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>x+1</math>, and we therefore use the quotient rule for differentiation:
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{x^{2}+1}{x+1} \right)^{\prime }=\frac{\left( x^{2}+1 \right)^{\prime }\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot \left( x+1 \right)^{\prime }}{\left( x+1 \right)^{2}} \\
		+	& \\
		+	& =\frac{2x\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot 1}{\left( x+1 \right)^{2}} \\
		+	& \\
		+	& =\frac{2x^{2}+2x-x^{2}-1}{\left( x+1 \right)^{2}}=\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}} \\
		+	\end{align}</math>
		+
		+
		+	NOTE: it is possible to rewrite the numerator by completing the square,
		+
		+
		+	<math>x^{2}+2x-1=\left( x+1 \right)^{2}-1^{2}-1=\left( x+1 \right)^{2}-2</math>
		+
		+
		+	and then the answer can be written as
		+
		+
		+
		+	<math>\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}}=\frac{\left( x+1 \right)^{2}-2}{\left( x+1 \right)^{2}}=1-\frac{2}{\left( x+1 \right)^{2}}</math>

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

Tekbot — Thu, 21 Aug 2008 07:51:04 GMT

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

← Nächstältere Version		Version vom 07:51, 21. Aug. 2008
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-	<center> [[~~Bild~~:1_2_1c.gif]] </center>	+	<center> [[Image:1_2_1c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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Oskar — Fri, 04 Apr 2008 08:32:34 GMT

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