A) e
B) \[\frac{1}{e}\]
C) \[{{e}^{2}}\]
D) \[\frac{1}{{{e}^{2}}}\]
Correct Answer: B
Solution :
[b] \[\because f(x)=lo{{g}_{x}}(logx)=\frac{{{\log }_{e}}(\log x)}{\log _{e}^{x}}\] \[f'(x)={{\frac{{{\log }_{e}}x.\frac{1}{{{\log }_{e}}x}\times \frac{1}{x}-{{\log }_{e}}(\log x).\frac{1}{x}}{{{(\log x)}^{2}}}}^{2}}\] \[\left[ \frac{d}{dx}\left( \frac{u}{v} \right)=\frac{v.du}{dx}-\frac{u.du}{dx} \right]=\frac{\frac{1}{x}-\log (\log x).\frac{1}{x}}{{{(\log x)}^{2}}}\] \[f'{{(x)}_{at\,x=e}}=\frac{\frac{1}{e}-\log 1.\frac{1}{e}}{{{({{\log }_{e}}e)}^{2}}}=1/e\] \[\left[ \therefore \log a=0\And {{\log }_{e}}e=1 \right]\]
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