${\text{Consider cot}}^{–\text{1}}\left(\frac{\frac{\sqrt{3}}{2}\mathrm{cosx}+\frac{1}{2}\mathrm{sinx}}{\frac{1}{2}\mathrm{sinx}-\frac{\sqrt{3}}{2}\mathrm{sinx}}\right)$

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

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

$\{\begin{array}{c}\frac{\mathrm{\pi }}{2}-(\mathrm{x}+\frac{\mathrm{\pi }}{3})=(\frac{\mathrm{\pi }}{6}-\mathrm{x});0<\mathrm{x}<\frac{\mathrm{\pi }}{6}\\ \frac{\mathrm{\pi }}{2}-((\mathrm{x}-\frac{\mathrm{\pi }}{3})-\mathrm{\pi })=(\frac{7\mathrm{\pi }}{6}-\mathrm{x});\frac{\mathrm{\pi }}{6}<\mathrm{x}<\frac{\mathrm{\pi }}{2}\end{array}$

$\therefore 2\mathrm{y}=\{\begin{array}{c}{(\frac{\mathrm{\pi }}{6}-\mathrm{x})}^2;0<\mathrm{x}<\frac{\mathrm{\pi }}{6}\\ {(\frac{7\mathrm{\pi }}{6}-\mathrm{x})}^2;\frac{\mathrm{\pi }}{6}<\mathrm{x}<\frac{\mathrm{\pi }}{2}\end{array}$

$\therefore 2\frac{\mathrm{dy}}{\mathrm{dx}}=\{\begin{array}{c}2(\frac{\mathrm{\pi }}{6}-\mathrm{x}).(-1);0<\mathrm{x}<\frac{\mathrm{\pi }}{6}\\ 2(\frac{7\mathrm{\pi }}{6}-\mathrm{x}).(-1);\frac{\mathrm{\pi }}{6}<\mathrm{x}<\frac{\mathrm{\pi }}{2}\end{array}$