Question

If $f\left(x\right)={\log }_5{\log }_{3}x$, then $f^{\prime }\left(e\right)$ is equal to

• A. $e{\log }_{e}5$
• B. $e{\log }_{e}3$
• C. $\frac{1}{e{\log }_{e}5}$
• D. $\frac{1}{e{\log }_{e}3}$

Answer 1

The correct option is C $\frac{1}{e{\log }_{e}5}$

Given, $f\left(x\right)={\log }_5{\log }_{3}x$

$⟹f\left(x\right)=\frac{\log {\log }_{3}x}{\log 5}$

$=\frac{1}{\log 5}\times \log {\log }_{3}x$

$=\frac{1}{\log 5}\times \log \left(\frac{\log x}{\log 3}\right)$

$=\frac{1}{\log 5}\times \left[\log \right(\log x)-\log (\log 3\left)\right]$

$\therefore f^{\prime }\left(x\right)=\frac{1}{\log 5}[\frac{1}{\log x}\cdot \frac{1}{x}-0]$

$⟹f^{\prime }\left(x\right)=\frac{1}{x\log x\log 5}$
$\therefore f^{\prime }\left(e\right)=\frac{1}{e\log e\log 5}=\frac{1}{e{\log }_{e}5}$

Hence, the answer is $\frac{1}{e{\log }_{e}5}$