Lösung 1.1:2c

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We differentiate term by term:
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<center> [[Image:1_1_2c.gif]] </center>
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<math>\begin{align}
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& {f}'\left( x \right)=\frac{d}{dx}\left( e^{x}-\ln x \right) \\
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& =\frac{d}{dx}e^{x}-\frac{d}{dx}\ln x=e^{x}-\frac{1}{x} \\
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\end{align}</math>
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NOTE: because
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<math>\text{ln }x</math>
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is not defined for
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<math>x\le 0</math>
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we assume implicitly that
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<math>x>0</math>.

Version vom 11:22, 10. Okt. 2008

We differentiate term by term:


\displaystyle \begin{align} & {f}'\left( x \right)=\frac{d}{dx}\left( e^{x}-\ln x \right) \\ & =\frac{d}{dx}e^{x}-\frac{d}{dx}\ln x=e^{x}-\frac{1}{x} \\ \end{align}


NOTE: because \displaystyle \text{ln }x is not defined for \displaystyle x\le 0 we assume implicitly that \displaystyle x>0.

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