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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{{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
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{{Displayed math||<math>\begin{align}
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{{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]

Version vom 12:50, 10. Mär. 2009

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}

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