Question

The derivative of the function \[\cot^{- 1} \left| \left( \cos 2 x \right)^{1/2} \right| \text{ at } x = \pi/6 \text{ is }\] ______ .

Options

• (2/3)^1/2

• (1/3)^1/2

• 3^1/2

• 6^1/2

Answer 1

(2/3)^1/2

^{\[\text{ We have, y } = \cot^{- 1} \left( \sqrt{\cos 2x} \right)\]}

\[\Rightarrow \frac{dy}{dx} = \frac{- 1}{1 + \cos 2x}\frac{d}{dx}\sqrt{\cos 2x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 1}{2 \cos^2 x} \times \frac{1}{2\sqrt{\cos 2x}}\frac{d}{dx}\left( \cos 2x \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 1}{2 \cos^2 x} \times \frac{1}{2\sqrt{\cos 2x}} \times - 2\sin 2x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin2x}{\cos^2 x \times 2\sqrt{\cos2x}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2 \sin x \cos x}{\cos^2 x \times 2\sqrt{\cos2x}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\tan x}{\sqrt{\cos2x}}\]
\[\text {So, at x } = \frac{\pi}{6},\text{ we get }\]
\[ \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{6}} = \frac{\tan\left( \frac{\pi}{6} \right)}{\sqrt{\cos2\left( \frac{\pi}{6} \right)}} = \frac{\left( \frac{1}{\sqrt{3}} \right)}{\sqrt{\frac{1}{2}}} = \left( \frac{2}{3} \right)^\frac{1}{2}\]