Question

If $f(x) = \log_{x^2} (\log \,x)$, then $f'(x)$ at $x = e$ is

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

Answer 1

The Correct Option is D

$f\left(x\right) = \log_{x^{2}}\left(\log x\right) $
$\Rightarrow \:\: f\left(x\right) =\frac{1}{2} \left[\log_{x}\left(\log x\right) \right] \left[\because \:\: \log_{a^{n}} b=\frac{1}{n} \log_{a^{b}} \right]$
$\Rightarrow f\left(x\right) =\frac{1}{2} \left[\frac{\log \left(\log x\right) }{\log x}\right] \left(\because \:\: \log_{a} b= \frac{\log b}{\log a } \right)$
$\Rightarrow f'\left(x\right) =\frac{1}{2} \left[\frac{\log\left(x\right) \frac{d}{dx}\left[\log \left(\log \left(x\right)\right)\right]-\log \left(\log x\right) \frac{d}{dx}\log x}{\left(\log x\right)^{2}}\right]$
$ =\frac{1}{2} \left[\frac{\log\left(x\right) \frac{1}{\log x}\times\frac{1}{x} -\log\left(\log x\right) \frac{1}{x}}{\left(\log x\right)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\frac{1}{x} -\frac{\log\left(\log \left(x\right)\right)}{x}}{\left(\log x\right)^{2}}\right]$
$ \Rightarrow f'\left(x\right) =\frac{1}{2}\left[\frac{1 -\log\left(\log \left(x\right)\right)}{x \, \times \,\left(\log x\right)^{2}}\right]$
$ \Rightarrow f'\left(e\right) =\frac{1}{2}\left[\frac{1 -\log\left(\log \left(e\right)\right)}{e \left(\log e\right)^{2}}\right]$
$ = \frac{1}{2} \left[ \frac{1 - 0}{e} \right] = \frac{1}{2e}$