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}} $
$=\frac{1-\ln \{(\ln x)\}}{x\{\ln (x)\}^{2}}$
$\Rightarrow f'(e) =\frac{1-\ln \{\ln (e)\}}{e\{\ln (e)\}^{2}}$
$=\frac{1-\ln (1)}{e}=\frac{1}{e}$