Lösung 1.2:3a - Versionsgeschichte

Sekretariat1 um 10:26, 20. Aug. 2009

Sekretariat1 — Thu, 20 Aug 2009 10:26:53 GMT

Martin um 11:30, 19. Apr. 2009

Martin — Sun, 19 Apr 2009 11:30:52 GMT

← Nächstältere Version		Version vom 11:30, 19. Apr. 2009
Zeile 1:		Zeile 1:
-	~~There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm~~,	+	Als erster Schritt berechnen wir die Ableitung der äußeren Logarithmusfunktion,

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

-	~~We can carry out the differentiation of~~ <math>\sqrt{x}+\sqrt{x+1}</math> ~~on the right-hand side term by term to obtain~~	+	Wir leiten den Ausdruck <math>\sqrt{x}+\sqrt{x+1}</math> Term für Term ab, und erhalten

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

-	~~and it remains then only to differentiate~~ <math>\sqrt{x}</math>~~, which we do directly, and~~ <math>\sqrt{x+1}</math> ~~(which has a simple inner derivative)~~,	+	Danach leiten wir die Funktionen <math>\sqrt{x}</math> und <math>\sqrt{x+1}</math> direkt ab,

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

-	~~If we rewrite the expression inside the square brackets using a common denominator, we get~~	+	Schreiben wir die Brüche mit gemeinsamen Nenner erhalten wir

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

-	~~and we can then eliminate the factor~~ <math>\sqrt{x+1}+\sqrt{x}</math> ~~from the numerator and denominator to get~~	+	Wir kürzen den Bruch mit <math>\sqrt{x+1}+\sqrt{x}</math> und erhalten

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

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

Tekbot — Wed, 11 Mar 2009 10:05:35 GMT

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

← Nächstältere Version

Version vom 10:05, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 12:53:54 GMT

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

Tek um 11:41, 15. Okt. 2008

Tek — Wed, 15 Oct 2008 11:41:43 GMT

← Nächstältere Version		Version vom 11:41, 15. Okt. 2008
Zeile 1:		Zeile 1:
-	There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:	+	There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm,

		+	{{Displayed math\|\|<math>\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr) = {}\rlap{\frac{1}{\bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)'\,\textrm{.}}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}

-	<math>~~\frac{d}{dx}\ln \left( \left\{ \left.~~ \sqrt{x}+\sqrt{x+1~~} \right\} \right. \right)=\frac{1}{\left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right.}\centerdot \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)^{\prime~~ }</math>	+	We can carry out the differentiation of <math>\sqrt{x}+\sqrt{x+1}</math> on the right-hand side term by term to obtain

		+	{{Displayed math\|\|<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{} = {}\rlap{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \bigl[ (\sqrt{x})' + (\sqrt{x+1})'\bigr]}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}

-	~~We can carry out the differentiation of~~	+	and it remains then only to differentiate <math>\sqrt{x}</math>, which we do directly, and <math>\sqrt{x+1}</math> (which has a simple inner derivative),
-	~~<math>\sqrt{x}+\sqrt{x+1}</math>~~	+
-	~~on the right-hand side term by term to obtain~~	+
-		+
-		+
-	~~<math>\quad =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \left( \sqrt{x} \right)^{\prime }+\left( \sqrt{x+1} \right)^{\prime } \right]</math>~~	+
-		+
-		+
-	and it remains then only to differentiate	+
-	<math>\sqrt{x}</math>,which we do directly, and	+
-	<math>\sqrt{x+1}</math>	+
-	which a simple inner derivative)	+
-		+
-		+
-	~~<math></math>~~	+
-		+
-	~~<math>\begin{align}~~	+
-	~~& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot \left( x+1 \right)^{\prime } \right] \\~~	+
-	~~& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot 1 \right] \\~~	+
-	~~\end{align}</math>~~	+
-		+

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

	If we rewrite the expression inside the square brackets using a common denominator, we get		If we rewrite the expression inside the square brackets using a common denominator, we get

		+	{{Displayed math\|\|<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
		+	= {}\rlap{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \Bigr]\,,}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}

-	~~<math>=\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \right]</math>,~~	+	and we can then eliminate the factor <math>\sqrt{x+1}+\sqrt{x}</math> from the numerator and denominator to get
-		+
-	and we can then eliminate the factor	+
-	<math>\sqrt{x+1}+\sqrt{x}</math>	+
-	from the numerator and denominator to get	+
-		+

-	<math>=\frac{1}{2\sqrt{x}\sqrt{x+1}}</math>	+	{{Displayed math\|\|<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
		+	= {}\rlap{\frac{1}{2\sqrt{x}\sqrt{x+1}}\,\textrm{.}}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}

Ian um 14:17, 11. Okt. 2008

Ian — Sat, 11 Oct 2008 14:17:19 GMT

← Nächstältere Version		Version vom 14:17, 11. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:
-	<~~center~~> ~~[[Image:1_2_3a~~-1(2)~~.gif]~~] </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\frac{d}{dx}\ln \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)=\frac{1}{\left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right.}\centerdot \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)^{\prime }</math>
-	~~<center>~~ [~~[Image:1_2_3a-~~2(2)~~.gif~~]] </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	We can carry out the differentiation of
		+	<math>\sqrt{x}+\sqrt{x+1}</math>
		+	on the right-hand side term by term to obtain
		+
		+
		+	<math>\quad =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \left( \sqrt{x} \right)^{\prime }+\left( \sqrt{x+1} \right)^{\prime } \right]</math>
		+
		+
		+	and it remains then only to differentiate
		+	<math>\sqrt{x}</math>,which we do directly, and
		+	<math>\sqrt{x+1}</math>
		+	which a simple inner derivative)
		+
		+
		+	<math></math>
		+
		+	<math>\begin{align}
		+	& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot \left( x+1 \right)^{\prime } \right] \\
		+	& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot 1 \right] \\
		+	\end{align}</math>
		+
		+
		+
		+	If we rewrite the expression inside the square brackets using a common denominator, we get
		+
		+
		+	<math>=\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \right]</math>,
		+
		+	and we can then eliminate the factor
		+	<math>\sqrt{x+1}+\sqrt{x}</math>
		+	from the numerator and denominator to get
		+
		+
		+	<math>=\frac{1}{2\sqrt{x}\sqrt{x+1}}</math>

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

Tekbot — Tue, 16 Sep 2008 14:26:01 GMT

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

← Nächstältere Version

Version vom 14:26, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:52:44 GMT

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_3a-1(2).gif]] </center>	+	<center> [[Image:1_2_3a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_3a-2(2).gif]] </center>	+	<center> [[Image:1_2_3a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:1_2_3a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:1_2_3a-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Fri, 04 Apr 2008 08:40:02 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_3a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_3a-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:1_2_3a-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:1_2_3a-2(2).gif]] </center>
{{NAVCONTENT_STOP}}

← Nächstältere Version		Version vom 10:26, 20. Aug. 2009
Zeile 1:		Zeile 1:
-	Als ~~erster~~ Schritt berechnen wir die Ableitung der äußeren Logarithmusfunktion,	+	Als ersten Schritt berechnen wir die Ableitung der äußeren Logarithmusfunktion.

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

-	Wir leiten den Ausdruck <math>\sqrt{x}+\sqrt{x+1}</math> Term für Term ab, und erhalten	+	Wir leiten den Ausdruck <math>\sqrt{x}+\sqrt{x+1}</math> Term für Term ab und erhalten

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

-	Danach leiten wir die Funktionen <math>\sqrt{x}</math> und <math>\sqrt{x+1}</math> direkt ab,	+	Danach leiten wir die Funktionen <math>\sqrt{x}</math> und <math>\sqrt{x+1}</math> direkt ab.

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

Zeile 18:		Zeile 18:

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

	Wir kürzen den Bruch mit <math>\sqrt{x+1}+\sqrt{x}</math> und erhalten		Wir kürzen den Bruch mit <math>\sqrt{x+1}+\sqrt{x}</math> und erhalten

← Nächstältere Version		Version vom 12:53, 10. Mär. 2009
Zeile 1:		Zeile 1:
	There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm,		There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm,

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

	We can carry out the differentiation of <math>\sqrt{x}+\sqrt{x+1}</math> on the right-hand side term by term to obtain		We can carry out the differentiation of <math>\sqrt{x}+\sqrt{x+1}</math> on the right-hand side term by term to obtain

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

	and it remains then only to differentiate <math>\sqrt{x}</math>, which we do directly, and <math>\sqrt{x+1}</math> (which has a simple inner derivative),		and it remains then only to differentiate <math>\sqrt{x}</math>, which we do directly, and <math>\sqrt{x+1}</math> (which has a simple inner derivative),

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}		\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
	&= \frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]\\[5pt]		&= \frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]\\[5pt]
Zeile 17:		Zeile 17:
	If we rewrite the expression inside the square brackets using a common denominator, we get		If we rewrite the expression inside the square brackets using a common denominator, we get

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

	and we can then eliminate the factor <math>\sqrt{x+1}+\sqrt{x}</math> from the numerator and denominator to get		and we can then eliminate the factor <math>\sqrt{x+1}+\sqrt{x}</math> from the numerator and denominator to get

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