Question

If $f\left(x\right)={\cos }^{-1}\left\{\frac{1-{\left({\log }_{e}x\right)}^2}{1+{\left({\log }_{e}x\right)}^2}\right\}$, then $f^{\prime }\left(e\right)$

• A. Does not exist
• B. Is equal to $\frac{2}{e}$
• C. Is equal to $\frac{1}{e}$
• D. Is equal to $1$

Answer 1

Answer: (C)

Is equal to $\frac{1}{e}$

We have
$f\left(x\right)={\cos }^{-1}\left\{\frac{1-{\left({\log }_{e}x\right)}^2}{1+{\left({\log }_{e}x\right)}^2}\right\}$
$=2{\tan }^{-1}\left({\log }_{e}x\right)[\because {\log }_{e}x>0]$ (in the $nbd$ of $x=e$)
$\Rightarrow f^{\prime }\left(x\right)=\frac{2}{1+{\left({\log }_{e}x\right)}^2}.\frac{1}{x}$
$\Rightarrow f^{\prime }\left(e\right)=\frac{2}{1+{\left({\log }_{e}e\right)}^2}.\frac{1}{e}$
$=\frac{2}{1+1}.\frac{1}{e}$
$=\frac{1}{e}$