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

The derivative of the function $\cot ^{-1}[\cos 2 x]^{1 / 2}$ at $x=\pi / 6$ is
  1. $\left(\frac{1}{3}\right)^{1 / 2}$
  2. $\left(\frac{2}{3}\right)^{1 / 2}$
  3. $\left(\frac{3}{2}\right)^{1 / 2}$
  4. $(3)^{1 / 2}$

Solution

$\begin{aligned} & f(x)=\cot ^{-1}\left[(\cot 2 x)^{\frac{1}{2}}\right]=\cot ^{-1}(\sqrt{\cos 2 x}) \\ & \therefore \quad f^{\prime}(x)=\frac{-1}{1+(\sqrt{\cos 2 x})^2} \times \frac{d}{d x}(\sqrt{\cos 2 x}) \\ & =\frac{-1}{1+\cos 2 x} \times \frac{1}{2 \sqrt{\cos 2 x}} \times(-2 \sin 2 x)=\frac{2 \sin 2 x}{2(1+\cos 2 x) \sqrt{\cos 2 x}} \\ & \therefore \quad\left[f^{\prime}(x)\right]_{x=\frac{\pi}{6}}=\frac{\sin \left(\frac{\pi}{3}\right)}{\left(1+\cos \frac{\pi}{3}\right) \sqrt{\cos \frac{\pi}{3}}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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