The value of derivative of the function cot^-1(cos 2x)^1/2 at x = (π)…

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

Let \(y = \cot^{-1}\left(\sqrt{\cos 2x}\right)\). By the chain rule, with \(\frac{d}{du}\cot^{-1}u = -\frac{1}{1 + u^2}\):

    \(\displaystyle \frac{dy}{dx} = -\frac{1}{1 + \cos 2x}\cdot\frac{d}{dx}\sqrt{\cos 2x}\)

    \(\displaystyle \frac{d}{dx}\sqrt{\cos 2x} = \frac{-2\sin 2x}{2\sqrt{\cos 2x}} = -\frac{\sin 2x}{\sqrt{\cos 2x}}\)

So

    \(\displaystyle \frac{dy}{dx} = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}\)

At \(x = \frac{\pi}{6}\), \(2x = \frac{\pi}{3}\):

    \(\displaystyle \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2},\quad \cos\frac{\pi}{3} = \frac{1}{2}\)

    \(\displaystyle \frac{dy}{dx} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}\cdot\frac{1}{\sqrt{2}}}\)

    \(\displaystyle = \frac{\sqrt{3}\cdot\sqrt{2}}{3} = \frac{\sqrt{6}}{3}\)

    \(\displaystyle = \sqrt{\frac{6}{9}} = \left(\frac{2}{3}\right)^{\frac{1}{2}}\)