JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Differentiation

Recognise the argument as a tangent: $\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}=\frac{\sin \left(x+\frac{\pi }{3}\right)}{\cos \left(x+\frac{\pi }{3}\right)}=\tan \left(x+\frac{\pi }{3}\right)$, so ${\cot }^{-1}$ of it is $\frac{\pi }{2}-{\tan }^{-1}\tan \left(x+\frac{\pi }{3}\right)$. For $0<x<\frac{\pi }{6}$ the angle $x+\frac{\pi }{3}$ lies between $\frac{\pi }{3}$ and $\frac{\pi }{2}$, so the inverse tangent returns $x+\frac{\pi }{3}$ and the whole expression is $\frac{\pi }{6}-x$. Hence $2y={\left(\frac{\pi }{6}-x\right)}^2$, and differentiating gives $2\frac{dy}{dx}=2\left(\frac{\pi }{6}-x\right)(-1)$, so $\frac{dy}{dx}=x-\frac{\pi }{6}$.