Solution 1.2:1e

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The quotient rule gives
The quotient rule gives
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\Bigl(\frac{x}{\ln x}\Bigr)'
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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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&= \frac{(x)'\cdot \ln x - x\cdot (\ln x)'}{(\ln x)^2}\\[5pt]
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& \\
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&= \frac{1\cdot\ln x - x\cdot\dfrac{1}{x}}{(\ln x)^2}\\[5pt]
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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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&= \frac{\ln x-1}{(\ln x)^2}\\[5pt]
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\end{align}</math>
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&= \frac{1}{\ln x} - \frac{1}{(\ln x)^2}\,\textrm{.}
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\end{align}</math>}}

Current revision

The quotient rule gives

\displaystyle \begin{align}

\Bigl(\frac{x}{\ln x}\Bigr)' &= \frac{(x)'\cdot \ln x - x\cdot (\ln x)'}{(\ln x)^2}\\[5pt] &= \frac{1\cdot\ln x - x\cdot\dfrac{1}{x}}{(\ln x)^2}\\[5pt] &= \frac{\ln x-1}{(\ln x)^2}\\[5pt] &= \frac{1}{\ln x} - \frac{1}{(\ln x)^2}\,\textrm{.} \end{align}

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