Question

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

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

Answer 1

Answer: (D)

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