Answer
$y'=-\dfrac{\pi(\cos\pi x)\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})](\sin\sqrt{\sin\pi x})}{\sqrt{\sin\pi x}}$
Work Step by Step
$y=\sin^{2}(\cos\sqrt{\sin\pi x})$
Start the differentiation process by using the chain rule:
$y'=2\sin(\cos\sqrt{\sin\pi x})[\sin(\cos\sqrt{\sin\pi x})]'=...$
Apply the chain rule one more time to evaluate the indicated derivative:
$...=2\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})][\cos\sqrt{\sin\pi x}]'=...$
Once again, apply the chain rule to evaluate the indicated derivative:
$...=-2\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})](\sin\sqrt{\sin\pi x})(\sqrt{\sin\pi x})'=...$
Apply the chain rule again to evaluate the indicated derivative:
$...=-2\Big(\dfrac{1}{2}\Big)\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})](\sin\sqrt{\sin\pi x})(\sin\pi x)^{-1/2}(\sin\pi x)'=...$
Evaluate the remaining derivative and simplify the expression:
$...=-\dfrac{\pi(\cos\pi x)\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})](\sin\sqrt{\sin\pi x})}{\sqrt{\sin\pi x}}$
