Question

If $f\left(x\right)={\cos }^{-1}\left[\frac{1-(logx{)}^2}{1+(logx{)}^2}\right]$, then the value of $f^{\prime }\left(e\right)$ is equal to

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

Answer 1

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

Given,

$f\left(x\right)=co{s}^{-1}\left[\frac{1-(logx{)}^2}{1+(logx{)}^2}\right]$

$f\left(x\right)=2ta{n}^{-1}\left(logx\right)$

$⟹f^{\prime }\left(x\right)=2\frac{1}{1+(logx{)}^2}\times \frac{1}{x}$

$⟹f^{\prime }\left(e\right)=\frac{2}{1+(loge{)}^2}\times \frac{1}{e}$

$⟹f^{\prime }\left(e\right)=\frac{1}{e}$