Question

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

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

Answer 1

The correct option is D $0$

$f\left(x\right)={\log }_{x^2}\left(\ln x\right)$

$=\frac{\log \left(\ln x\right)}{\log x^2}$

$\Rightarrow f\left(x\right)=\frac{1}{2\log x}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}\frac{-1}{(\log x{)}^2}\frac{1}{x}$

$f^{\prime }\left(x\right)atx=\log e$

$f^{\prime }\left(\log e\right)=\frac{1}{2}\frac{-1}{\left(\log \right(\log e){)}^2}\frac{1}{\log e}$

$f^{\prime }\left(\log e\right)=0$