Question

The derivative of ${\cot }^{-1}\left(\sqrt{\cos 2x}\right)$ at $x=\frac{\pi }{6}$ is -

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{\frac{1}{3}}$
• C. $\sqrt{3}$
• D. $\sqrt{6}$

Answer 1

The correct option is A $\sqrt{\frac{2}{3}}$

$y={\cot }^{-1}\left[\sqrt{\cos 2x}\right]$
$\frac{dy}{dx}=\frac{-1}{1+\cos 2x}\times \frac{1}{2\sqrt{\cos 2x}}\times -\sin 2x\times 2$

${\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{6}}=\frac{+1}{(1+\frac{1}{2})}\times \frac{1}{2\times \frac{1}{\sqrt{2}}}\frac{\sqrt{3}}{2}\times 2$

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