Lösung 1.2:4a - Versionsgeschichte

Sekretariat1 um 11:03, 20. Aug. 2009

Sekretariat1 — Thu, 20 Aug 2009 11:03:11 GMT

← Nächstältere Version		Version vom 11:03, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir beginnen damit die Funktion einmal abzuleiten. Durch die Quotientenregel erhalten wir	+	Wir beginnen damit, die Funktion einmal abzuleiten. Durch die Quotientenregel erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	Die Ableitung <math>\bigl(\sqrt{1-x^2}\bigr)'</math> erhalten wir durch die Kettenregel,	+	Die Ableitung <math>\bigl(\sqrt{1-x^2}\bigr)'</math> erhalten wir durch die Kettenregel.

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

-	Wir vereinfachen die Ableitung so weit ~~wir~~ möglich, sodass wir die zweite Ableitung einfacher berechnen können,	+	Wir vereinfachen die Ableitung so weit wie möglich, sodass wir die zweite Ableitung einfacher berechnen können.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 22:		Zeile 22:
	&= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]		&= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
	&= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]		&= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
-	&= \frac{1}{(1-x^2)^{3/2}}\,\textrm{.}	+	&= \frac{1}{(1-x^2)^{3/2}}\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

Zeile 34:		Zeile 34:
	&= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-5/2}\cdot (-2x)\\[5pt]		&= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-5/2}\cdot (-2x)\\[5pt]
	&= 3x\bigl(1-x^2\bigr)^{-5/2}\\[5pt]		&= 3x\bigl(1-x^2\bigr)^{-5/2}\\[5pt]
-	&= \frac{3x}{\bigl(1-x^2\bigr)^{5/2}}\,\textrm{.}	+	&= \frac{3x}{\bigl(1-x^2\bigr)^{5/2}}\,\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Martin um 12:34, 19. Apr. 2009

Martin — Sun, 19 Apr 2009 12:34:37 GMT

← Nächstältere Version		Version vom 12:34, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We are to differentiate the expression two times, so we start by differentiating once~~. ~~The quotient rule gives~~	+	Wir beginnen damit die Funktion einmal abzuleiten. Durch die Quotientenregel erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We determine the derivative~~ <math>\bigl(\sqrt{1-x^2}\bigr)'</math> ~~by using the chain rule~~	+	Die Ableitung <math>\bigl(\sqrt{1-x^2}\bigr)'</math> erhalten wir durch die Kettenregel,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 15:		Zeile 15:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We simplify the result as far as possible,~~ so ~~as to make the second differentiation easier~~,	+	Wir vereinfachen die Ableitung so weit wir möglich, sodass wir die zweite Ableitung einfacher berechnen können,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 25:		Zeile 25:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The second derivative is~~	+	Die zweite Ableitung ist

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

Tekbot: Solution 1.2:4a moved to Lösung 1.2:4a: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:07:35 GMT

Solution 1.2:4a moved to Lösung 1.2:4a: Robot: moved page

← Nächstältere Version

Version vom 10:07, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 12:54:54 GMT

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

← Nächstältere Version		Version vom 12:54, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives		We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}		\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
	&= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]		&= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 9:		Zeile 9:
	We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule		We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}		\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
	&= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]		&= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
Zeile 17:		Zeile 17:
	We simplify the result as far as possible, so as to make the second differentiation easier,		We simplify the result as far as possible, so as to make the second differentiation easier,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}		\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
	&= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]		&= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 27:		Zeile 27:
	The second derivative is		The second derivative is

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}		\frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
	&= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]		&= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]

Tek um 13:47, 15. Okt. 2008

Tek — Wed, 15 Oct 2008 13:47:55 GMT

← Nächstältere Version		Version vom 13:47, 15. Okt. 2008
Zeile 1:		Zeile 1:
	We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives		We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives

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

-	<math>\~~begin{align}~~	+	We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
-	~~& \frac{d}{dx}\frac{x}{\sqrt{1-x^{2}}}=\frac{\left~~( ~~x \right)^{\prime }~~\sqrt{1-x^{2}~~}-x~~\~~left( \sqrt{1-x^{2}} \right~~)~~^{\prime }}{\left( \sqrt{1-x^{2}} \right)^{2}} \\~~	+
-	~~& \\~~	+
-	~~& =\frac{1\centerdot \sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{1-x^{2}} \\~~	+
-	~~\end{align}~~</math>	+

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

-	We ~~determine~~ the ~~derivative~~	+	We simplify the result as far as possible, so as to make the second differentiation easier,
-	~~<math>\left( \sqrt{1-x^{2}} \right)^{\prime }</math>~~	+
-	~~by using~~ the ~~chain rule~~	+

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

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

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~We simplify the result as far as possible, so as to make the second differentiation easier:~~	+	\frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
-		+	&= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]
-		+	&= \frac{d}{dx}\,\bigl(1-x^2\bigr)^{-3/2}\\[5pt]
-	<math>\begin{align}	+	&= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-3/2-1}\cdot\bigl(1-x^2\bigr)'\\[5pt]
-	~~& =~~\frac{~~\sqrt{1-x~~^{2}~~}+\frac~~{x^{2}}{\~~sqrt{1-x^{2}}}}{1-x^{2}} \\~~	+	&= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-5/2}\cdot (-2x)\\[5pt]
-	~~& \\~~	+	&= 3x\bigl(1-x^2\bigr)^{-5/2}\\[5pt]
-	~~& =~~\frac{~~\frac{\left( \sqrt{1-~~x~~^{2}} \right)^{2}~~}{\sqrt{1-x^{2}}~~}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\~~	+	&= \frac{3x}{\bigl(1-x^2\bigr)^{5/2}}\,\textrm{.}
-	~~& \\~~	+	\end{align}</math>}}
-	& =\frac{~~\frac{1-x^{2~~}~~+x^~~{2}}{\~~sqrt{1-x^{2}}}}{1-x^{2}} \\~~	+
-	~~& \\~~	+
-	~~& =~~\frac{1}{~~\left~~( 1-x^{2~~} \right~~)^{{3}/{2~~}\;~~}} \\	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~The second derivative is:~~	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	& ~~\frac{d^{^{2}}}{dx^{^{2}}}\frac{x}{\sqrt{1-x^{2}}}~~=\frac{d}{dx}\~~frac{1}{~~\~~left~~( 1-x^{2} \~~right~~)^{{3}/{2~~}\;}~~} \\	+
-	~~& \\~~	+
-	& =~~\frac{d}{dx}\left( 1~~-~~x^{2}~~ \~~right)^{-{3}/{2}\;}=-\frac~~{3}{2}\~~left~~( 1-x^{2} \~~right~~)^{-~~\frac{~~3}{2}-1}\~~centerdot~~ \~~left~~( 1-x^{2} \~~right~~)~~^{\prime } \\~~	+
-	& \\	+
-	& =-\~~frac~~{3}{2}\~~left~~( 1-x^{2} \~~right~~)^{{-5}/{2}\~~;}\centerdot \left~~( -2x \~~right)~~=3x\~~left~~( 1-x^{2} \~~right~~)^{{-5}/{2}~~\;} \\~~	+
-	& \\	+
-	& =\frac{3x}{\~~left~~( 1-x^{2} \~~right~~)^{{5}/{2~~}\;~~}} \\	+
-	\end{align}</math>	+

Ian um 14:04, 12. Okt. 2008

Ian — Sun, 12 Oct 2008 14:04:59 GMT

← Nächstältere Version		Version vom 14:04, 12. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
-	~~<center> [[Image:1_2_4a~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\begin{align}
-	<~~center~~> ~~[[Image~~:~~1_2_4a~~-2(2)~~.gif]]~~ </~~center~~>	+	& \frac{d}{dx}\frac{x}{\sqrt{1-x^{2}}}=\frac{\left( x \right)^{\prime }\sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{\left( \sqrt{1-x^{2}} \right)^{2}} \\
-	{{~~NAVCONTENT_STOP~~}}	+	& \\
		+	& =\frac{1\centerdot \sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{1-x^{2}} \\
		+	\end{align}</math>
		+
		+
		+	We determine the derivative
		+	<math>\left( \sqrt{1-x^{2}} \right)^{\prime }</math>
		+	by using the chain rule
		+
		+
		+	<math>\begin{align}
		+	& =\frac{\sqrt{1-x^{2}}-x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( 1-x^{2} \right)^{\prime }}{1-x^{2}} \\
		+	& \\
		+	& =\frac{\sqrt{1-x^{2}}-x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right)}{1-x^{2}} \\
		+	\end{align}</math>
		+
		+
		+	We simplify the result as far as possible, so as to make the second differentiation easier:
		+
		+
		+	<math>\begin{align}
		+	& =\frac{\sqrt{1-x^{2}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\
		+	& \\
		+	& =\frac{\frac{\left( \sqrt{1-x^{2}} \right)^{2}}{\sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\
		+	& \\
		+	& =\frac{\frac{1-x^{2}+x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\
		+	& \\
		+	& =\frac{1}{\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
		+	\end{align}</math>
		+
		+
		+	The second derivative is:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d^{^{2}}}{dx^{^{2}}}\frac{x}{\sqrt{1-x^{2}}}=\frac{d}{dx}\frac{1}{\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
		+	& \\
		+	& =\frac{d}{dx}\left( 1-x^{2} \right)^{-{3}/{2}\;}=-\frac{3}{2}\left( 1-x^{2} \right)^{-\frac{3}{2}-1}\centerdot \left( 1-x^{2} \right)^{\prime } \\
		+	& \\
		+	& =-\frac{3}{2}\left( 1-x^{2} \right)^{{-5}/{2}\;}\centerdot \left( -2x \right)=3x\left( 1-x^{2} \right)^{{-5}/{2}\;} \\
		+	& \\
		+	& =\frac{3x}{\left( 1-x^{2} \right)^{{5}/{2}\;}} \\
		+	\end{align}</math>

Tekbot: Lösning 1.2:4a moved to Solution 1.2:4a: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:28:01 GMT

Lösning 1.2:4a moved to Solution 1.2:4a: Robot: moved page

← Nächstältere Version

Version vom 14:28, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:53:44 GMT

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

← Nächstältere Version		Version vom 07:53, 21. Aug. 2008
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_4a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:1_2_4a-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Fri, 04 Apr 2008 08:43:38 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_4a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_4a-2(2).gif </center> {{NAVCONTENT_STOP}}

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