Lösung 1.2:1e

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The quotient rule gives
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<math>\begin{align}
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& \left( \frac{x}{\ln x} \right)^{\prime }=\frac{\left( x \right)^{\prime }\centerdot \ln x-x\centerdot \left( \ln x \right)^{\prime }}{\left( \ln x \right)^{2}} \\
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& \\
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& =\frac{1\centerdot \ln x-x\centerdot \frac{1}{x}}{\left( \ln x \right)^{2}}=\frac{\ln x-1}{\left( \ln x \right)^{2}}=\frac{1}{\ln x}-\frac{1}{\left( \ln x \right)^{2}} \\
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\end{align}</math>

Version vom 15:39, 12. Sep. 2008

The quotient rule gives


\displaystyle \begin{align} & \left( \frac{x}{\ln x} \right)^{\prime }=\frac{\left( x \right)^{\prime }\centerdot \ln x-x\centerdot \left( \ln x \right)^{\prime }}{\left( \ln x \right)^{2}} \\ & \\ & =\frac{1\centerdot \ln x-x\centerdot \frac{1}{x}}{\left( \ln x \right)^{2}}=\frac{\ln x-1}{\left( \ln x \right)^{2}}=\frac{1}{\ln x}-\frac{1}{\left( \ln x \right)^{2}} \\ \end{align}

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