If 2y = ( ^ - 1 ( 3 x + x x - 3 x )) ^ 2 x ∈ (0, (π)/(2)) , then...

2y = {\left(\cot^{- 1}\dfrac{\sqrt{3}\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)}^{2}$</p><p>$= {\left(\cot^{- 1}\left(\dfrac{\sqrt{3} + \tan x}{1 - \sqrt{3}\tan x}\right)\right)}^{2}$</p><p>$= {\left(\cot^{- 1}\tan \left(\dfrac{\pi}{3} + x\right)\right)}^{2}$</p><p>$= {\left(\cot^{- 1}\cot \left(\dfrac{\pi}{2} - \left(\dfrac{\pi}{3} + x\right)\right)\right)}^{2}$</p><p>$⇒ 2y = \left\{$$\begin{array}{ll}{\left(\dfrac{\pi}{6} - x\right)}^{2}, x \in \left(0, \dfrac{\pi}{6}\right) \\ {\left(\pi + \dfrac{\pi}{6} - x\right)}^{2}, x \in \left(\dfrac{\pi}{6}, \dfrac{\pi}{2}\right)\end{array}$$\right)

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

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

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