Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 40

Find $y'$ of $y = \sin^2 \left( \cos \sqrt{\sin 'pi x}\right)$


$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left[ \sin^2 \left( \cos \sqrt{\sin 'pi x}\right) \right]\\
\\
y' &= \frac{d}{dx} \left[ \sin^2 \left( \cos \sqrt{\sin 'pi x}\right) \right]^2\\
\\
y' &= 2 \sin \left( \cos \sqrt{\sin \pi x} \right) \frac{d}{dx} [\sin(\cos\sqrt{\sin \pi x})]\\
\\
y' &= 2 \sin (\cos\sqrt{\sin \pi x}) \cdot (\cos \sqrt{\sin \pi x}) \frac{d}{dx} (\cos\sqrt{\sin \pi x})\\
\\
y' &= 2 \sin (\cos\sqrt{\sin \pi x}) \cdot (\cos \sqrt{\sin \pi x}) \cdot (-\sin\sqrt{\sin \pi x}) \frac{d}{dx} (\sqrt{\sin \pi x})\\
\\
y' &= 2 \sin (\cos\sqrt{\sin \pi x}) \cdot (\cos \sqrt{\sin \pi x}) \cdot (-\sin\sqrt{\sin \pi x}) \frac{d}{dx} (\sin \pi x)^{\frac{1}{2}}\\
\\
y' &= 2 \sin (\cos\sqrt{\sin \pi x}) \cdot (\cos \sqrt{\sin \pi x}) \cdot (-\sin\sqrt{\sin \pi x}) \cdot \left( \frac{1}{2}\right)(\sin \pi x)^{\frac{-1}{2}} \frac{d}{dx} ( \sin \pi x)\\
\\
y' &= 2 \sin (\cos\sqrt{\sin \pi x}) \cdot (\cos \sqrt{\sin \pi x}) \cdot (-\sin\sqrt{\sin \pi x}) \cdot \left( \frac{1}{2}\right)(\sin \pi x)^{\frac{-1}{2}} \cdot (\cos \pi x) \frac{d}{dx} (\pi x)\\
\\
y' &= \cancel{2} \sin (\cos\sqrt{\sin \pi x}) \cdot (\cos \sqrt{\sin \pi x}) \cdot (-\sin\sqrt{\sin \pi x}) \cdot \left( \frac{1}{\cancel{2}}\right)(\sin \pi x)^{\frac{-1}{2}} \cdot (\cos \pi x)(\pi)\\
\\
y' &= \frac{-\pi \sin(\cos\sqrt{\sin \pi x})\cos(\cos\sqrt{\sin \pi x}) \sin \sqrt{\sin \pi x} \cos \pi x}{(\sin \pi x)^{\frac{1}{2}}} \qquad \text{or} \qquad y' = \frac{-\pi \sin(\cos\sqrt{\sin \pi x})\cos(\cos\sqrt{\sin \pi x}) \sin \sqrt{\sin \pi x} \cos \pi x}{\sqrt{\sin \pi x}}
\end{aligned}
\end{equation}
$