Answer
$$ y'=\frac{3\sin x}{2\sqrt{4-3\cos x}}.$$
Work Step by Step
Since $ y=\sqrt{4-3\cos x}=(4-3\cos x)^{1/2}$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\cos x)'=-\sin x $, the derivative $ y'$ is given by
$$ y'=\frac{1}{2}(4-3\cos x)^{-1/2}(4-3\cos x)'
\\=\frac{-3(-\sin x)}{2\sqrt{4-3\cos x}}=\frac{3\sin x}{2\sqrt{4-3\cos x}}.$$
