Lösung 1.1:2a - Versionsgeschichte

Ombtut3 um 12:51, 5. Sep. 2009

Ombtut3 — Sat, 05 Sep 2009 12:51:53 GMT

← Nächstältere Version		Version vom 12:51, 5. Sep. 2009
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	&= 2x-3\,\textrm{.}		&= 2x-3\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}
		+
		+	Alternativer Lösungsweg: [[1.1:2a_alternativ_1\|Limes]]

Sekretariat1 — Wed, 19 Aug 2009 10:50:34 GMT

← Nächstältere Version		Version vom 10:50, 19. Aug. 2009
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	{{Abgesetzte Formel\|\|<math>\frac{d}{dx} \left( f(x) + g(x) \right) = \frac{d}{dx} \left( f(x) \right) + \frac{d}{dx} \left( g(x) \right)</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx} \left( f(x) + g(x) \right) = \frac{d}{dx} \left( f(x) \right) + \frac{d}{dx} \left( g(x) \right)</math>}}

-	~~Erhalten~~ wir	+	erhalten wir

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

Martin — Thu, 09 Apr 2009 09:56:15 GMT

← Nächstältere Version		Version vom 09:56, 9. Apr. 2009
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	und		und

-	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\\left( f(x) + g(x) \right) = \frac{d}{dx}\\left( f(x) \right) + \frac{d}{dx}\\left( g(x) \right)</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx} \left( f(x) + g(x) \right) = \frac{d}{dx} \left( f(x) \right) + \frac{d}{dx} \left( g(x) \right)</math>}}

	Erhalten wir		Erhalten wir

Martin — Thu, 09 Apr 2009 09:55:42 GMT

← Nächstältere Version		Version vom 09:55, 9. Apr. 2009
Zeile 1:		Zeile 1:
-	~~By using the rule for differentiation~~	+	Durch die Regeln

	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}}

-	~~and the fact that the expression can be differentiated term by term~~ ~~and that constant factors can be taken outside the differentiation, we obtain~~	+	und
		+
		+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\\left( f(x) + g(x) \right) = \frac{d}{dx}\\left( f(x) \right) + \frac{d}{dx}\\left( g(x) \right)</math>}}
		+
		+	Erhalten wir

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

Tekbot — Wed, 11 Mar 2009 09:58:35 GMT

← Nächstältere Version

Version vom 09:58, 11. Mär. 2009

Tekbot — Tue, 10 Mar 2009 12:50:24 GMT

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

← Nächstältere Version		Version vom 12:50, 10. Mär. 2009
Zeile 1:		Zeile 1:
	By using the rule for differentiation		By using the rule for differentiation

-	{{~~Displayed math~~\|\|<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}}

	and the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain		and the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain

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

Tek — Tue, 14 Oct 2008 11:45:30 GMT

← Nächstältere Version		Version vom 11:45, 14. Okt. 2008
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	By using the rule for differentiation		By using the rule for differentiation

		+	{{Displayed math\|\|<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}}

-	~~<math>\frac{d}{dx}x^{n}=nx^{n-1}</math>~~	+	and the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain~~	+	f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^2-3x+1\bigr)\\[5pt]
-		+	&= \frac{d}{dx}\,x^2 - 3\frac{d}{dx}\,x^1 + \frac{d}{dx}\,1\\[5pt]
-		+	&= 2x^{2-1} - 3\cdot 1x^{1-1} + 0\\[5pt]
-	<math>\begin{align}	+	&= 2x-3\,\textrm{.}
-	~~& {~~f}'\~~left~~( x ~~\right~~)=\frac{d}{dx}\~~left~~( x^{2}-3x+1 \~~right~~)=\frac{d}{dx}x^{2}-3\frac{d}{dx}x^{1}+\frac{d}{dx}1 \\	+	\end{align}</math>}}
-	& =2x^{2-1}-3\~~centerdot~~ 1x^{1-1}+0=2x-3 \\	+
-	\end{align}</math>	+

Ian — Fri, 10 Oct 2008 10:55:35 GMT

← Nächstältere Version		Version vom 10:55, 10. Okt. 2008
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-	{{~~NAVCONTENT_START~~}}	+	By using the rule for differentiation
-	<~~center~~> ~~[[Image:1_1_2a.gif]]~~ <~~/center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\frac{d}{dx}x^{n}=nx^{n-1}</math>
		+
		+
		+	the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain
		+
		+
		+	<math>\begin{align}
		+	& {f}'\left( x \right)=\frac{d}{dx}\left( x^{2}-3x+1 \right)=\frac{d}{dx}x^{2}-3\frac{d}{dx}x^{1}+\frac{d}{dx}1 \\
		+	& =2x^{2-1}-3\centerdot 1x^{1-1}+0=2x-3 \\
		+	\end{align}</math>

Tekbot — Tue, 16 Sep 2008 14:19:01 GMT

← Nächstältere Version

Version vom 14:19, 16. Sep. 2008

Tekbot — Thu, 21 Aug 2008 07:49:14 GMT

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