The value of derivative of the function $\cot^{-1}\{(\cos2x)^{1/2}\}$ at $x =\frac{\pi}{6}$ is

$\left(\frac{2}{3}\right)^{\frac{1}{2}}$

The correct answer is Option (1) → $\left(\frac{2}{3}\right)^{\frac{1}{2}}$

Let:

$f(x) = \cot^{-1} \left( \sqrt{ \cos 2x } \right)$

Step 1: Differentiate using chain rule

$f'(x) = \frac{d}{dx} \left[ \cot^{-1} \left( \sqrt{ \cos 2x } \right) \right]$

$= \frac{-1}{1 + \left( \sqrt{ \cos 2x } \right)^2} \cdot \frac{d}{dx} \left( \sqrt{ \cos 2x } \right)$

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

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

Step 2: Evaluate at $x = \frac{\pi}{6}$

$\cos 2x = \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$

$\sin 2x = \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$

Now substitute:

$f'\left( \frac{\pi}{6} \right) = \frac{ \frac{\sqrt{3}}{2} }{ \left(1 + \frac{1}{2} \right) \cdot \sqrt{ \frac{1}{2} } }$

$= \frac{ \frac{\sqrt{3}}{2} }{ \frac{3}{2} \cdot \frac{1}{\sqrt{2}} }$

$= \frac{ \sqrt{3} }{3 } \cdot \sqrt{2} = \frac{ \sqrt{6} }{3 }$

Now simplify carefully

$\frac{ \sqrt{6} }{3 } = \sqrt{ \frac{6}{9} } = \sqrt{ \frac{2}{3} }$