If $f(x)=\cot ^{-1}(\cos 2 x)^{1 / 2}$, then $f'\left(\frac{\pi}{6}\right)$ is :

$\sqrt{\frac{2}{3}}$

$f'(x)= \frac{-1}{1+\cos 2 x} . \frac{1}{2}(\cos 2 x)^{-1 / 2}(-\sin 2 x . 2)$

$= \frac{\sin 2 x}{\sqrt{\cos 2 x(1+\cos 2 x)}}$

$\Rightarrow f'\left(\frac{\pi}{6}\right)=\frac{\sin \frac{\pi}{3}}{\sqrt{\cos \frac{\pi}{3}}\left(1+\cos \frac{\pi}{3}\right)}$

$=\frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{1}{2}}\left(1+\frac{1}{2}\right)}=\sqrt{\frac{2}{3}}$

Hence (3) is correct answer.