Lösung 1.1:2a

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By using the rule for differentiation
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<center> [[Image:1_1_2a.gif]] </center>
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<math>\frac{d}{dx}x^{n}=nx^{n-1}</math>
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the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain
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<math>\begin{align}
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& {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 \\
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& =2x^{2-1}-3\centerdot 1x^{1-1}+0=2x-3 \\
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\end{align}</math>

Version vom 10:55, 10. Okt. 2008

By using the rule for differentiation


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


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}'\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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:2a“