From Förberedande kurs i matematik 2

Revision as of 15:13, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_1_2a.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (11:45, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	By using the rule for differentiation
-	<center> [[Bild:1_1_2a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed 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}
		+	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]
		+	&= 2x-3\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

By using the rule for differentiation

\displaystyle \frac{d}{dx}\,x^{n}=nx^{n-1}

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

\displaystyle \begin{align} 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] &= 2x-3\,\textrm{.} \end{align}