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

The function is $f(x) = \cot^{-1}\{(\cos 2x)^{1/2}\}$

Let $y = \cot^{-1}\{(\cos 2x)^{1/2}\}$

Let $u = (\cos 2x)^{1/2}$

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The derivative formula for inverse cotangent:

$\frac{d}{dx}[\cot^{-1}(u)] = -\frac{1}{1+u^2} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = -\frac{1}{1+[(\cos 2x)^{1/2}]^2} \cdot \frac{d}{dx}[(\cos 2x)^{1/2}]$

$\frac{dy}{dx} = -\frac{1}{1+\cos 2x} \cdot \frac{d}{dx}[(\cos 2x)^{1/2}]$

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Finding $\frac{d}{dx}[(\cos 2x)^{1/2}]$:

$\frac{d}{dx}[(\cos 2x)^{1/2}] = \frac{1}{2}(\cos 2x)^{-1/2} \cdot \frac{d}{dx}(\cos 2x)$

$= \frac{1}{2(\cos 2x)^{1/2}} \cdot (-\sin 2x) \cdot 2$

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

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Combining:

$\frac{dy}{dx} = -\frac{1}{1+\cos 2x} \cdot \left(-\frac{\sin 2x}{(\cos 2x)^{1/2}}\right)$

$\frac{dy}{dx} = \frac{\sin 2x}{(1+\cos 2x)(\cos 2x)^{1/2}}$

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At $x = \frac{\pi}{6}$, we have $2x = \frac{\pi}{3}$

$\cos \frac{\pi}{3} = \frac{1}{2}$

$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

$\frac{dy}{dx}\bigg|_{x=\pi/6} = \frac{\frac{\sqrt{3}}{2}}{(1+\frac{1}{2})(\frac{1}{2})^{1/2}}$

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

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

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

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

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

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

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

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

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