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