Question

If $f\left(x\right)={\log }_{x^2}\left({\log }_{e}x\right)$, then $f^{\prime }\left(x\right)$ at $x=e$ is

• A. $1$
• B. $\frac{1}{e}$
• C. $\frac{1}{2e}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{2e}$

Given, $f\left(x\right)={\log }_{x^2}\left({\log }_{e}x\right)=\frac{1}{2}{\log }_x\left({\log }_{e}x\right)$
$\Rightarrow f\left(x\right)=\frac{1}{2}\frac{{\log }_e{\log }_{e}x}{{\log }_{e}x}$
On differentiating both sides with respect to $x$, we get
$f^{\prime }\left(x\right)=\frac{1}{2}\frac{{\log }_{e}x\left(\frac{1}{x{\log }_{e}x}\right)-{\log }_e{\log }_{e}x\times \frac{1}{x}}{{\left({\log }_{e}x\right)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}\frac{[\frac{1}{x}-\frac{1}{x}{\log }_e{\log }_{e}x]}{{\left({\log }_{e}x\right)}^2}$

At $x=e$, $f^{\prime }\left(e\right)=\frac{1}{2}\frac{[\frac{1}{e}-\frac{1}{e}{\log }_{e}1]}{1^2}$

$\Rightarrow f^{\prime }\left(e\right)=\frac{1}{2}[\frac{1}{e}-0]$ ....... $[\because {\log }_{e}1=0]$
$\Rightarrow f^{\prime }\left(e\right)=\frac{1}{2e}$