Lösung 1.2:2c - Versionsgeschichte

Sekretariat1 um 07:44, 20. Aug. 2009

2009-08-20T07:44:41Z

← Nächstältere Version		Version vom 07:44, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir sehen dass die äußere Funktion	+	Wir sehen, dass die äußere Funktion

-	{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{}</math>}}

-	ist, die wir zuerst ableiten und dann mit der inneren Ableitung von <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math> multiplizieren,	+	ist, die wir zuerst ableiten und dann mit der inneren Ableitung von <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math> multiplizieren

-	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,.</math>}}

-	Wir haben hier folgende Regeln benutzt,	+	Wir haben hier folgende Regeln benutzt

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}

Martin um 20:57, 18. Apr. 2009

2009-04-18T20:57:42Z

← Nächstältere Version		Version vom 20:57, 18. Apr. 2009
Zeile 1:		Zeile 1:
-	~~When we see this expression, we should think "square root of something",~~	+	Wir sehen dass die äußere Funktion

	{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}

-	~~and in order to differentiate it~~, ~~we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression~~ <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,	+	ist, die wir zuerst ableiten und dann mit der inneren Ableitung von <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math> multiplizieren,

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}

-	~~where we have used the differentiation rule~~	+	Wir haben hier folgende Regeln benutzt,

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}

-	~~Thus, we obtain~~	+	Daher erhalten wir

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}

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

2009-03-11T10:04:15Z

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

← Nächstältere Version

Version vom 10:04, 11. Mär. 2009

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

2009-03-10T12:53:14Z

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

← Nächstältere Version		Version vom 12:53, 10. Mär. 2009
Zeile 1:		Zeile 1:
	When we see this expression, we should think "square root of something",		When we see this expression, we should think "square root of something",

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

	and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,		and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,

-	{{~~Displayed math~~\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}

	where we have used the differentiation rule		where we have used the differentiation rule

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

	Thus, we obtain		Thus, we obtain

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

Tek um 08:15, 15. Okt. 2008

2008-10-15T08:15:01Z

← Nächstältere Version		Version vom 08:15, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	When we see this expression, we should think "root of something",	+	When we see this expression, we should think "square root of something",

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

-	<math>\~~sqrt~~{\~~left\~~{ \~~left. {~~} ~~\right\~~} \~~right.}~~</math>	+	and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,

-		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}
-	and in order to differentiate it, we should first differentiate the outer function , "the root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression	+
-		+
-	~~<math>\left\~~{ ~~\left.~~ {~~} \right\} \right.=\cos x</~~math>,	+
-		+
-		+
-	<math>\frac{d}{dx}\sqrt{\~~left\~~{ ~~\left.~~ \cos x ~~\right\~~} ~~\right.~~}=\frac{1}{2\sqrt{\~~left\{ \left~~. \cos x ~~\right\~~} ~~\right.~~}}\~~centerdot~~ \~~left~~( \~~left\~~{ ~~\left.~~ \cos x ~~\right\~~} \~~right. \right~~)^{\~~prime }~~</math>,	+

	where we have used the differentiation rule		where we have used the differentiation rule

-		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
-	<math>\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2~~}\;~~}=\~~frac~~{1}{2}x^~~{\frac~~{1}{2}-1}=\~~frac~~{1}{2}x^{-~~\frac{~~1}{2}}=\frac{1}{2\sqrt{x}}</math>.	+

	Thus, we obtain		Thus, we obtain

-		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}
-	<math>\frac{d}{dx}\sqrt{\cos x}=\frac{1}{2\sqrt{\cos x}}\~~centerdot \left~~( -\sin x ~~\right~~)=-\frac{\sin x}{2\sqrt{\cos x}}</math>	+

Ian um 13:06, 11. Okt. 2008

2008-10-11T13:06:24Z

← Nächstältere Version		Version vom 13:06, 11. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	When we see this expression, we should think "root of something",
-	<~~center~~> ~~[[Image:1_2_2c~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\sqrt{\left\{ \left. {} \right\} \right.}</math>
		+
		+
		+	and in order to differentiate it, we should first differentiate the outer function , "the root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression
		+
		+	<math>\left\{ \left. {} \right\} \right.=\cos x</math>,
		+
		+
		+	<math>\frac{d}{dx}\sqrt{\left\{ \left. \cos x \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \cos x \right\} \right.}}\centerdot \left( \left\{ \left. \cos x \right\} \right. \right)^{\prime }</math>,
		+
		+	where we have used the differentiation rule
		+
		+
		+	<math>\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}</math>.
		+
		+	Thus, we obtain
		+
		+
		+	<math>\frac{d}{dx}\sqrt{\cos x}=\frac{1}{2\sqrt{\cos x}}\centerdot \left( -\sin x \right)=-\frac{\sin x}{2\sqrt{\cos x}}</math>

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

2008-09-16T14:24:41Z

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

← Nächstältere Version

Version vom 14:24, 16. Sep. 2008

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

2008-08-21T07:52:04Z

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

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2008-04-04T08:36:39Z

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