Ответ:

Используем правило \((\operatorname{ctg}u)'=-u'/\sin^2u\):

\(y'=-4\cdot\dfrac{1/2}{\sin^2(\dfrac{x}{2}+\dfrac{\pi}{6})}=-\dfrac{2}{\sin^2(\dfrac{x}{2}+\dfrac{\pi}{6})}.\)

Ответ: \(y'=-\dfrac{2}{\sin^2(\dfrac{x}{2}+\dfrac{\pi}{6})}\).