Lösung 1.2:3c - Versionsgeschichte

Sekretariat1 um 10:44, 20. Aug. 2009

2009-08-20T10:44:46Z

Martin um 12:24, 19. Apr. 2009

2009-04-19T12:24:42Z

← Nächstältere Version		Version vom 12:24, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We can write the expression as~~	+	Wir schreiben useren Ausdruck wie

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

-	~~and then we see that we have~~ "~~something raised to~~ -1"~~, which can be differentiated one step by using the chain rule,~~	+	Und sehen dass die äußere Funktion "irgendetwas hoch -1" ist. Verwenden wir die Kettenregel erhalten wir die Ableitung

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 12:		Zeile 12:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The next step is to differentiate the product~~ <math>x\cdot\sqrt{1-x^2}</math> ~~using the product rule~~,	+	Die Aleitung von <math>x\cdot\sqrt{1-x^2}</math> erhalten wir durch die Faktorregel,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 20:		Zeile 20:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The expression~~ <math>\sqrt{1-x^2}</math> ~~is of the type "square root of something", so we use the chain rule to differentiate,~~	+	Wir verwenden wieder die Kettenregel um <math>\sqrt{1-x^2}</math> abzuleiten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 29:		Zeile 29:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We write the expression on the right over a common denominator,~~	+	Schreiben wir den Ausdruck mit gemeinsamen Nenner erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 39:		Zeile 39:


-	~~Note~~: ~~When we make simplifications of the form <math>\bigl~~(\sqrt{1-x^2} \bigr)^2 = 1-x^2</math>, ~~we assume that both sides are well defined~~ (~~i.e.~~ in ~~this case that~~ <math>x</math> ~~lies between~~ -1 ~~and~~ 1).	+	Hinweis: Wenn wir vereinfachungen wir (\sqrt{1-x^2} \bigr)^2 = 1-x^2</math> machen, nehmen wir an dass beide Seiten definiert sind (in diesen Fall das <math>x</math> zwischen -1 und 1 liegt).

Tekbot: Solution 1.2:3c moved to Lösung 1.2:3c: Robot: moved page

2009-03-11T10:06:15Z

Solution 1.2:3c moved to Lösung 1.2:3c: Robot: moved page

← Nächstältere Version

Version vom 10:06, 11. Mär. 2009

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

2009-03-10T12:54:14Z

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

Tek um 12:55, 15. Okt. 2008

2008-10-15T12:55:49Z

← Nächstältere Version		Version vom 12:55, 15. Okt. 2008
Zeile 1:		Zeile 1:
	We can write the expression as		We can write the expression as

		+	{{Displayed math\|\|<math>\frac{1}{x\sqrt{1-x^{2}}} = \bigl(x\sqrt{1-x^2}\,\bigr)^{-1}\,\textrm{,}</math>}}

-	~~<math>\frac{1}{x\sqrt{1-x^{2}}}=\left( x\sqrt{1-x^{2}} \right)^{~~-1~~}</math>~~,	+	and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,

-	~~and then we see that we have "something raised to~~	+	{{Displayed math\|\|<math>\begin{align}
-	<math>-\~~text~~{1}</math>~~", which can be differentiated one step by using the chain rule:~~	+	\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}
		+	&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
		+	&= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt]
		+	&= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{.}
		+	\end{align}</math>}}

		+	The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule,

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

		+	The expression <math>\sqrt{1-x^2}</math> is of the type "square root of something", so we use the chain rule to differentiate,

-	~~The next step is to differentiate the product~~	+	{{Displayed math\|\|<math>\begin{align}
-	<math>x\~~centerdot~~ \sqrt{1-x^{2}}</math>	+	\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
-	~~using the product rule:~~	+	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]
		+	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot ( -2x)\Bigr)\\[5pt]
		+	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{.}
		+	\end{align}</math>}}

		+	We write the expression on the right over a common denominator,

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


-	~~The expression~~	+	Note: When we make simplifications of the form <math>\bigl(\sqrt{1-x^2} \bigr)^2 = 1-x^2</math>, we assume that both sides are well defined (i.e. in this case that <math>x</math> lies between -1 and 1).
-	~~<math>\sqrt{1-x^{2}}</math>~~	+
-	~~is of the type "root of something", so we use the chain rule to differentiate,~~	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\left( 1-x^{2} \right)^{\prime } \right) \\~~	+
-	~~& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right) \right) \\~~	+
-	~~& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}} \right) \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~We write the expression on the right over a common denominator:~~	+
-		+
-	~~<math>\begin{align}~~	+
-	~~& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{\left( \sqrt{1-x^{2}} \right)^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\~~	+
-	~~& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{1-x^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\~~	+
-	~~& =-\frac{1-2x^{2}}{x^{2}\left( 1-x^{2} \right)^{{3}/{2}\;}} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~NOTE~~: When we make simplifications of the form	+
-	<math>\~~left~~( \sqrt{1-x^{2}} \~~right~~)^{2}=1-x^{2}</math>, we assume that both sides are well defined (i.e. in this case that	+
-	<math>x</math>	+
-	lies between	+
-	~~<math>~~-~~\text{~~1~~}</math>~~	+
-	and	+
-	~~<math>\text{~~1~~}</math>~~	+
-	).	+

Ian um 11:57, 15. Okt. 2008

2008-10-15T11:57:10Z

← Nächstältere Version		Version vom 11:57, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We can write the expression as
-	<~~center~~> ~~[[Image~~:~~1_2_3c~~-1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\frac{1}{x\sqrt{1-x^{2}}}=\left( x\sqrt{1-x^{2}} \right)^{-1}</math>,
-	<~~center~~> ~~[[Image~~:~~1_2_3c~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	and then we see that we have "something raised to
		+	<math>-\text{1}</math>", which can be differentiated one step by using the chain rule:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)^{-1}=-1\centerdot \left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)\centerdot \left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)^{\prime } \\
		+	& =-\frac{1}{\left( x\sqrt{1-x^{2}} \right)^{2}}\centerdot \left( x\sqrt{1-x^{2}} \right)^{\prime } \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( x\sqrt{1-x^{2}} \right)^{\prime } \\
		+	\end{align}</math>
		+
		+
		+	The next step is to differentiate the product
		+	<math>x\centerdot \sqrt{1-x^{2}}</math>
		+	using the product rule:
		+
		+
		+	<math>\begin{align}
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \left( x \right)^{\prime }\sqrt{1-x^{2}}+x\left( \sqrt{1-x^{2}} \right)^{\prime } \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( 1\sqrt{1-x^{2}}+x\left( \sqrt{1-x^{2}} \right)^{\prime } \right) \\
		+	\end{align}</math>
		+
		+
		+	The expression
		+	<math>\sqrt{1-x^{2}}</math>
		+	is of the type "root of something", so we use the chain rule to differentiate,
		+
		+
		+	<math>\begin{align}
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\left( 1-x^{2} \right)^{\prime } \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right) \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}} \right) \\
		+	\end{align}</math>
		+
		+
		+	We write the expression on the right over a common denominator:
		+
		+	<math>\begin{align}
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{\left( \sqrt{1-x^{2}} \right)^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{1-x^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\
		+	& =-\frac{1-2x^{2}}{x^{2}\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
		+	\end{align}</math>
		+
		+
		+	NOTE: When we make simplifications of the form
		+	<math>\left( \sqrt{1-x^{2}} \right)^{2}=1-x^{2}</math>, we assume that both sides are well defined (i.e. in this case that
		+	<math>x</math>
		+	lies between
		+	<math>-\text{1}</math>
		+	and
		+	<math>\text{1}</math>
		+	).

Tekbot: Lösning 1.2:3c moved to Solution 1.2:3c: Robot: moved page

2008-09-16T14:26:41Z

Lösning 1.2:3c moved to Solution 1.2:3c: Robot: moved page

← Nächstältere Version

Version vom 14:26, 16. Sep. 2008

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

2008-08-21T07:53:04Z

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

← Nächstältere Version		Version vom 07:53, 21. Aug. 2008
Zeile 1:		Zeile 1:
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-	<center> [[~~Bild~~:1_2_3c-1(2).gif]] </center>	+	<center> [[Image:1_2_3c-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_3c-2(2).gif]] </center>	+	<center> [[Image:1_2_3c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_3c-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:1_2_3c-2(2).gif {{NAVCONTENT_STOP}}

2008-04-04T08:42:19Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_3c-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_3c-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:1_2_3c-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
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<center> [[Bild:1_2_3c-2(2).gif]] </center>
{{NAVCONTENT_STOP}}

← Nächstältere Version		Version vom 10:44, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir schreiben ~~useren~~ Ausdruck wie	+	Wir schreiben unseren Ausdruck wie folgt

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

-	~~Und~~ sehen dass die äußere Funktion "irgendetwas hoch -1" ist. Verwenden wir die Kettenregel erhalten wir die Ableitung	+	Wir sehen, dass die äußere Funktion "irgendetwas hoch -1" ist. Verwenden wir die Kettenregel, erhalten wir die Ableitung.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 9:		Zeile 9:
	&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]		&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
	&= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt]		&= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt]
-	&= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{.}	+	&= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

-	Die ~~Aleitung~~ von <math>x\cdot\sqrt{1-x^2}</math> erhalten wir durch die Faktorregel,	+	Die Ableitung von <math>x\cdot\sqrt{1-x^2}</math> erhalten wir durch die Faktorregel.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}		\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
	&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]		&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
-	&= -\frac{1}{x^2(1-x^2)}\cdot \Bigl(1\cdot\sqrt{1-x^2} + x\cdot (\sqrt{1-x^2})'\Bigr)\,\textrm{.}	+	&= -\frac{1}{x^2(1-x^2)}\cdot \Bigl(1\cdot\sqrt{1-x^2} + x\cdot (\sqrt{1-x^2})'\Bigr)\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

-	Wir verwenden wieder die Kettenregel um <math>\sqrt{1-x^2}</math> abzuleiten	+	Wir verwenden wieder die Kettenregel, um <math>\sqrt{1-x^2}</math> abzuleiten.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 26:		Zeile 26:
	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]		&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]
	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot ( -2x)\Bigr)\\[5pt]		&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot ( -2x)\Bigr)\\[5pt]
-	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{.}	+	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{}
	\end{align}</math>}}		\end{align}</math>}}

-	Schreiben wir den Ausdruck mit gemeinsamen Nenner erhalten wir	+	Schreiben wir den Ausdruck mit gemeinsamen Nenner, erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 39:		Zeile 39:


-	Hinweis: Wenn wir ~~vereinfachungen wir~~ (\sqrt{1-x^2} \bigr)^2 = 1-x^2</math> ~~machen~~, nehmen wir an dass beide Seiten definiert sind (in ~~diesen~~ Fall ~~das~~ <math>x</math> zwischen -1 und 1 liegt).	+	Hinweis: Wenn wir Vereinfachungen wie <math>(\sqrt{1-x^2} \bigr)^2 = 1-x^2</math> erstellen, nehmen wir an, dass beide Seiten definiert sind (in diesem Fall, dass <math>x</math> zwischen -1 und 1 liegt).

← Nächstältere Version		Version vom 12:54, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We can write the expression as		We can write the expression as

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

	and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,		and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}		\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}
	&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]		&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
Zeile 14:		Zeile 14:
	The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule,		The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}		\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
	&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]		&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
Zeile 22:		Zeile 22:
	The expression <math>\sqrt{1-x^2}</math> is of the type "square root of something", so we use the chain rule to differentiate,		The expression <math>\sqrt{1-x^2}</math> is of the type "square root of something", so we use the chain rule to differentiate,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}		\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
	&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]		&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]
Zeile 31:		Zeile 31:
	We write the expression on the right over a common denominator,		We write the expression on the right over a common denominator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}		\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
	&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \frac{\bigl(\sqrt{1-x^2}\bigr)^2-x^2}{\sqrt{1-x^2}}}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]		&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \frac{\bigl(\sqrt{1-x^2}\bigr)^2-x^2}{\sqrt{1-x^2}}}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]