Solution 1.2:1e
From Förberedande kurs i matematik 2
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- | {{ | + | The quotient rule gives |
- | + | ||
- | {{ | + | |
+ | <math>\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}</math> |
Revision as of 15:39, 12 September 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}