Q. If $y = \cot^{- 1}\ {(\cos{2x})}^{1/2}\ ,then\ the\ value\ of\ \frac{dy}{dx}\ at\ x = \frac{\pi}{6}\ will\ be\$
A) $\left( \frac{2}{3} \right)^{1/2}$
B) $\left( \frac{1}{3} \right)^{1/2}$
C) $(3)^{1/2}$
D) $(6)^{1/2}$
Explanation
(a)
Given, $y = \cot^{- 1}{(\cos{2x)}}^{1/2}$
$\Rightarrow \ \frac{dy}{dx}\ = \ \frac{- 1}{1 + \cos{2x}} \times \frac{1}{2\sqrt{\cos{2x}}} \times - 2\sin{2x}$
$= \ \frac{2\sin{x\cos x}}{2\cos^{2}{x\sqrt{\cos{2x}}}}$
$\Rightarrow \ \frac{dy}{dx} = \frac{\tan x}{\sqrt{\cos{2x}}}$
$\Rightarrow \ \ \ \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{6}} = \frac{1/\sqrt{3}}{\sqrt{1/2}} = \sqrt{\frac{2}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.