Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 45

Determine the derivative of the function $y = \cos \sqrt{\sin(\tan \pi x)}$


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