If 2 y = cot - 1 3 cos x + sin x cos x - 3 sin x 2   ∀ x ∈ 0 , π 2 , then d y d x…

$2y={\left({\cot }^{-1}\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)}^2$

$={\left({\cot }^{-1}\left(\frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}\right)\right)}^2$

$={\left({\cot }^{-1}\tan \left(\frac{\pi }{3}+x\right)\right)}^2$

$={\left({\cot }^{-1}\cot \left(\frac{\pi }{2}-\left(\frac{\pi }{3}+x\right)\right) \right)}^2$

$\Rightarrow 2y=\left\{\begin{array}{l}{\left(\frac{\pi }{6}-x\right)}^2, x\in \left(0,\frac{\pi }{6}\right)\\ {\left(\pi +\frac{\pi }{6}-x\right)}^2, x\in \left(\frac{\pi }{6},\frac{\pi }{2}\right)\end{array}\right.$
$$

$\Rightarrow \frac{2dy}{dx}=2\left(\frac{\pi }{6}-x\right).\left(-1\right)\Rightarrow \frac{dy}{dx}=x-\frac{\pi }{6}, x\in \left(0,\frac{\pi }{6}\right)$

And $\frac{dy}{dx}=x-\frac{7\pi }{6}, x\in \left(\frac{\pi }{6},\frac{\pi }{2}\right)$
Left Hand and Right Hand Derivatives are not same so function is non derivable at $x=\frac{\mathrm{\pi }}{6}$.

Hence, $\frac{dy}{dx}$ does not exist for all values in the given interval.