Вычислите производную в точке

Решите задачу:

$g(x)=4ctgx;\\ g'(-\frac{5\pi}{3})-?;\\ g'(x)=(4ctgx)'=4(ctgx)'=4(\frac{\cos x}{\sin x})'=4\frac{(\cos x)'\cdot\sin x-\cos x\cdot(\sin x)'}{\sin^2x}=\\ =4\frac{-\sin x\cdot\sinx-\cos x\cdot\cos x}{\sin^2x}=-4\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{-4}{\sin^2x};\\ g'(-\frac{2\pi}{3})=\frac{-4}{\sin^2(-\frac{2\pi}{3})}=\frac{-4}{(\frac{\sqrt{3}}{2})^2}=\frac{-4}{\frac{3}{4}}=-\frac{16}{3}=-5\frac{1}{3}$