## Calculus: Early Transcendentals 8th Edition

$y'=-\dfrac{\pi(\cos\pi x)\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})](\sin\sqrt{\sin\pi x})}{\sqrt{\sin\pi x}}$
$y=\sin^{2}(\cos\sqrt{\sin\pi x})$ 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: $...=-\dfrac{\pi(\cos\pi x)\sin(\cos\sqrt{\sin\pi x})[\cos(\cos\sqrt{\sin\pi x})](\sin\sqrt{\sin\pi x})}{\sqrt{\sin\pi x}}$