Question

If $f\left(x\right)=co{s}^{-1}\left[\frac{1-{\left(logx\right)}^2}{1+{\left(logx\right)}^2}\right]$, then $f^{\prime }\left(e\right)=...$

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

Answer 1

Answer: (A)

We have $f(x)={\cos }^{-1}\left[\frac{1-(\log x{)}^2}{1+(\log x{)}^2}\right]$
$f(x)=2{\tan }^{-1}(\log x)\left[\because {\cos }^{-1}\left(\frac{1-x^2}{.1+x^2}\right)=2{\tan }^{-1}x\right]$
$f^{\prime }(x)=\frac{2}{1+(\log x{)}^2}\times \frac{1}{x}$
$f^{\prime }(e)=\frac{2}{1+1}\times \frac{1}{e}=\frac{1}{e}$