Question

If $f(x)={\log }_x\{\ln (x)\}$, then $f^{\prime }(x)$ at $x=e$, is

• A. $e$
• B. $-e$
• C. $e^2$
• D. $e^{-1}$

Answer 1

Answer: (D)

We have, $f(x)={\log }_x\{\ln (x)\}=\frac{\ln \{\ln (x)\}}{\ln (x)}$
$\therefore f^{\prime }(x)=\frac{\ln (x)\cdot \frac{1}{\ln (x)}\cdot \frac{1}{x}-\ln \{\ln (x)\}\frac{1}{x}}{\{\ln (x){\}}^2}$
$=\frac{1-\ln \{(\ln x)\}}{x\{\ln (x){\}}^2}$
$\Rightarrow f^{\prime }(e)=\frac{1-\ln \{\ln (e)\}}{e\{\ln (e){\}}^2}$
$=\frac{1-\ln (1)}{e}=\frac{1}{e}$