Answer
$$y' =-\frac{1}{2}\pi\frac{\sin{\sqrt{\sin(\tan{\pi x})}}
\times \cos(\tan{\pi x})
\times \sec^2{\pi x}}{\sqrt{\sin(\tan{\pi x})}}$$
Work Step by Step
Given $y=\cos{\sqrt{\sin(\tan{\pi x})}}$.
$y'=\frac{d\cos{\sqrt{\sin(\tan{\pi x})}}}{dx}$
Using the chain rule:
$y'$
$=\frac{d\cos{\sqrt{\sin(\tan{\pi x})}}}{d\sqrt{\sin(\tan{\pi x})}}
\times \frac{d\sqrt{\sin(\tan{\pi x})}}{d\sin(\tan{\pi x})}
\times \frac{d\sin(\tan{\pi x})}{d\tan{\pi x}}
\times \frac{{d\tan{\pi x}}}{d\pi x}
\times \frac{d\pi x}{dx}$
$=-\sin{\sqrt{\sin(\tan{\pi x})}}
\times \frac{1}{2\sqrt{\sin(\tan{\pi x})}}
\times \cos(\tan{\pi x})
\times \sec^2{\pi x}
\times \pi$
Simplifying the above expression,
$y' =-\frac{1}{2}\pi\frac{\sin{\sqrt{\sin(\tan{\pi x})}}
\times \cos(\tan{\pi x})
\times \sec^2{\pi x}}{\sqrt{\sin(\tan{\pi x})}}$
