Question

If $f \left(x\right)=cos^{-1} \left[\frac{1-\left(log\,x\right)^{2}}{1+\left(log\,x\right)^{2}}\right]$, then $f'\left(e\right) = ...$

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

Answer 1

Answer: (A)

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