Question

If $f\left(x\right)={\log }_x\left(\ln x\right)$, then

• A. $f^{\prime }\left(e\right)=e$
• B. $f^{\prime }\left(e\right)=\frac{1}{e}$
• C. $f^{\prime }\left(e^2\right)=\frac{1}{e^2}$
• D. $f^{\prime }\left(e^2\right)=\frac{(1-\ln 2)}{(2e{)}^2}$

Answer 1

Answer: (B), (D)

The correct options are
B $f^{\prime }\left(e\right)=\frac{1}{e}$
D $f^{\prime }\left(e^2\right)=\frac{(1-\ln 2)}{(2e{)}^2}$

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

So,
$f^{\prime }\left(e\right)=\frac{1}{e}$ and
$f^{\prime }\left(e^2\right)=\frac{\frac{1}{e^2}[1-{\log }_{e}2]}{4}\Rightarrow f^{\prime }\left(e^2\right)=\frac{(1-{\log }_{e}2)}{(2e{)}^2}$