Lösung 1.1:2a

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By using the rule for differentiation
By using the rule for differentiation
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{{Displayed math||<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}}
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<math>\frac{d}{dx}x^{n}=nx^{n-1}</math>
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and 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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{{Displayed math||<math>\begin{align}
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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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f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^2-3x+1\bigr)\\[5pt]
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&= \frac{d}{dx}\,x^2 - 3\frac{d}{dx}\,x^1 + \frac{d}{dx}\,1\\[5pt]
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&= 2x^{2-1} - 3\cdot 1x^{1-1} + 0\\[5pt]
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<math>\begin{align}
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&= 2x-3\,\textrm{.}
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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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\end{align}</math>}}
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& =2x^{2-1}-3\centerdot 1x^{1-1}+0=2x-3 \\
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\end{align}</math>
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Version vom 11:45, 14. Okt. 2008

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}

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