Answer
$y'=\dfrac{-\pi(\cos{(\tan{\pi x})})(\sec^2{\pi x})(\sin{\sqrt{\sin{(\tan{\pi x})}}})}{2\sqrt{\sin{(\tan{\pi x})}}}$
Work Step by Step
Using the Chain Rule:
$u=\sqrt{\sin{(\tan{\pi x})}}=\sqrt{v}$
$v=\sin{(\tan{\pi x})}=\sin{z}$
$z=\tan{\pi x}$;
$\dfrac{dz}{dx}=\pi\sec^2{\pi x}$
$\dfrac{dv}{dx}=\dfrac{dv}{dz}\times \dfrac{dz}{dx}=(\cos{(\tan{\pi x})})(\pi\sec^2{\pi x})$
$\dfrac{du}{dx}=\dfrac{du}{dv}\times\dfrac{dv}{dx}$
$=(\dfrac{1}{2\sqrt{\sin{(\tan{\pi x})}}})((\cos{(\tan{\pi x})})(\pi\sec^2{\pi x}))$
$=\dfrac{(\cos{(\tan{\pi x})})(\pi\sec^2{\pi x})}{2\sqrt{\sin{(\tan{\pi x})}}}$
$\dfrac{dy}{du}=-\sin{u}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}$
$=(\dfrac{(\cos{(\tan{\pi x})})(\pi\sec^2{\pi x})}{2\sqrt{\sin{(\tan{\pi x})}}})(-\sin{\sqrt{\sin{(\tan{\pi x})}}})$
$=\dfrac{-\pi(\cos{(\tan{\pi x})})(\sec^2{\pi x})(\sin{\sqrt{\sin{(\tan{\pi x})}}})}{2\sqrt{\sin{(\tan{\pi x})}}}$
