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