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

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


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