Question

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

• A. $(2/3{)}^{1/2}$
• B. $(1/3{)}^{1/2}$
• C. $3^{1/2}$
• D. $6^{1/2}$

Answer 1

The correct option is A $(2/3{)}^{1/2}$

We have,

$y={\cot }^{-1}\left\{{\left(\cos 2x\right)}^{\frac{1}{2}}\right\}$

So,

$y={\cot }^{-1}\left(\sqrt{\cos 2x}\right)$

On differentiation with respect to x and we get,

$\frac{dy}{dx}=-\frac{1}{1+{\left(\sqrt{\cos 2x}\right)}^2}\frac{d}{dx}\sqrt{\cos 2x}$

$\frac{dy}{dx}=-\frac{1}{1+{\left(\sqrt{\cos 2x}\right)}^2}\frac{1}{2\sqrt{\cos 2x}}\frac{d}{dx}\cos 2x$

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

$\frac{dy}{dx}=\frac{1}{1+\cos 2x}\frac{\sin 2x}{\sqrt{\cos 2x}}$

At $x=\frac{\pi }{6}$

$\frac{dy}{dx}=\frac{1}{1+\cos \frac{2\pi }{6}}\frac{\sin 2\frac{\pi }{6}}{\sqrt{\cos \frac{2\pi }{6}}}$

$\frac{dy}{dx}=\frac{1}{1+\cos \frac{\pi }{3}}\frac{\sin \frac{\pi }{3}}{\sqrt{\cos \frac{\pi }{3}}}$

$\frac{dy}{dx}=\frac{1}{1+\frac{1}{2}}\frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{1}{2}}}$

$\frac{dy}{dx}=\frac{1}{\frac{3}{2}}\frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{1}{2}}}$

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

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

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

Hence, this is the answer.