Question

The derivative of the function ${\cot }^{-1}\left|{\left(\cos 2x\right)}^{1/2}\right|\mathrm{at}x=\pi /6\mathrm{is}$
(a) (2/3)^1/2
(b) (1/3)^1/2
(c) 3^1/2
(d) 6^1/2

Answer 1

(a) (2/3)^1/2
$\mathrm{We}\mathrm{have},y={\cot }^{-1}\left(\sqrt{\cos 2x}\right)$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{-1}{1+\cos 2x}\frac{d}{dx}\sqrt{\cos 2x} \\ \Rightarrow \frac{dy}{dx}=\frac{-1}{2{\cos }^{2}x}\times \frac{1}{2\sqrt{\cos 2x}}\frac{d}{dx}\left(\cos 2x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{-1}{2{\cos }^{2}x}\times \frac{1}{2\sqrt{\cos 2x}}\times -2\sin 2x \\ \Rightarrow \frac{dy}{dx}=\frac{\sin 2x}{{\cos }^{2}x\times 2\sqrt{\cos 2x}} \\ \Rightarrow \frac{dy}{dx}=\frac{2\sin x\cos x}{{\cos }^{2}x\times 2\sqrt{\cos 2x}} \\ \Rightarrow \frac{dy}{dx}=\frac{\tan x}{\sqrt{\cos 2x}} \\ \mathrm{So},\mathrm{at}x=\frac{\mathrm{\pi }}{6},\mathrm{we}\mathrm{get} \\ {\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{6}}=\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{\sqrt{\cos 2\left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}}=\frac{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}{\sqrt{{\displaystyle \frac{1}{2}}}}={\left(\frac{2}{3}\right)}^{\frac{1}{2}} \end{gathered}$$