Answer
$$y'=\frac{-\pi\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)\sec^2\pi x}{2\sqrt{\sin(\tan \pi x)}}$$
Work Step by Step
$$y'= \left(\cos\sqrt{\sin(\tan \pi x)}\right)'=-\sin\sqrt{\sin(\tan \pi x)}\cdot\left(\sqrt{\sin(\tan \pi x)}\right)'=-\sin\sqrt{\sin(\tan \pi x)}\frac{1}{2\sqrt{\sin(\tan \pi x)}}\cdot(\sin(\tan \pi x))'=\frac{-\sin\sqrt{\sin(\tan \pi x)}}{2\sqrt{\sin(\tan \pi x)}}\cdot \cos(\tan \pi x)\cdot(\tan \pi x)'=
\frac{-\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)}{2\sqrt{\sin(\tan \pi x)}}\frac{1}{\cos^2(\pi x)}(\pi x)'=
\frac{-\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)}{2\sqrt{\sin(\tan \pi x)}}\sec^2\pi x\cdot \pi=
\frac{-\pi\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)\sec^2\pi x}{2\sqrt{\sin(\tan \pi x)}}$$
