Question

If $f(x)=lo{g}_x$ { $ln(x)$ },then $f^{\prime }(e)$ is equal to

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

Answer 1

Answer: (A)

Given, $f(x)={\log }_x\{\log (x)\}$
$=\frac{\log \log x}{\log x}\left({\log }_{a}b=\frac{\log b}{\log a}\right)$
On differentiating w.r.t. ' $x$ ', we get
$f^{\prime }(x)=\frac{\log x\left[\frac{1}{\log x}\times \frac{1}{x}-\log \log x\times \frac{1}{x}\right]}{(\log x{)}^2}$
$=\frac{\left[\frac{1}{x\log x}-\frac{\log \log x}{x}\right]}{\log x}$
At $x=e$
$f^{\prime }(e)=\frac{\frac{1}{e\log e}-\frac{\log \log e}{e}}{\log e}$
$=\frac{\frac{1}{e}-0}{1}=\frac{1}{e}(\log \log e=\log 1=0)$