The derivative of the function cot ^{-1}[cos | vedclass

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(B) Let $f(x) = \cot^{-1}(\sqrt{\cos 2x})$.Using the chain rule,the derivative is $f'(x) = \frac{-1}{1 + (\sqrt{\cos 2x})^2} 	imes \frac{d}{dx}(\sqrt

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$\left(\frac{2}{3}\right)^{1 / 2}$

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(B) Let $f(x) = \cot^{-1}(\sqrt{\cos 2x})$.Using the chain rule,the derivative is $f'(x) = \frac{-1}{1 + (\sqrt{\cos 2x})^2} 	imes \frac{d}{dx}(\sqrt{\cos 2x})$.$f'(x) = \frac{-1}{1 + \cos 2x} 	imes \frac{1}{2\sqrt{\cos 2x}} 	imes (-\sin 2x 	imes 2)$.$f'(x) = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}$.At $x = \frac{\pi}{6}$,$2x = \frac{\pi}{3}$.$f'\left(\frac{\pi}{6}ight) = \frac{\sin(\pi/3)}{(1 + \cos(\pi/3))\sqrt{\cos(\pi/3)}}$.$f'\left(\frac{\pi}{6}ight) = \frac{\sqrt{3}/2}{(1 + 1/2)\sqrt{1/2}} = \frac{\sqrt{3}/2}{(3/2)(1/\sqrt{2})} = \frac{\sqrt{3}}{2} 	imes \frac{2}{3} 	imes \sqrt{2} = \frac{\sqrt{6}}{3} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}} = \left(\frac{2}{3}ight)^{1/2}$.

The derivative of the function $\cot ^{-1}[\cos 2 x]^{1 / 2}$ at $x=\pi / 6$ is

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