Question

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

• A. does not exist
• B. $\frac{2}{e}$
• C. $\frac{1}{e}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{e}$

Given : $f\left(x\right)={\cos }^{-1}\left[\frac{1-{\left(\log x\right)}^2}{1+{\left(\log x\right)}^2}\right]$
$\Rightarrow f\left(x\right)=2{\tan }^{-1}\left(\log x\right)$ ....... $[\because {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x]$

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

$=\frac{2}{(1+{\left(\log x\right)}^2)x}$

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

$=\frac{2}{(1+1))e}=\frac{2}{2e}=\frac{1}{e}$ ..... $[\because \log e=1]$
$\therefore f^{\prime }\left(e\right)=\frac{1}{e}$