## Calculus: Early Transcendentals 8th Edition

$$y' =-\frac{1}{2}\pi\frac{\sin{\sqrt{\sin(\tan{\pi x})}} \times \cos(\tan{\pi x}) \times \sec^2{\pi x}}{\sqrt{\sin(\tan{\pi x})}}$$
Given $y=\cos{\sqrt{\sin(\tan{\pi x})}}$. $y'=\frac{d\cos{\sqrt{\sin(\tan{\pi x})}}}{dx}$ Using the chain rule: $y'$ $=\frac{d\cos{\sqrt{\sin(\tan{\pi x})}}}{d\sqrt{\sin(\tan{\pi x})}} \times \frac{d\sqrt{\sin(\tan{\pi x})}}{d\sin(\tan{\pi x})} \times \frac{d\sin(\tan{\pi x})}{d\tan{\pi x}} \times \frac{{d\tan{\pi x}}}{d\pi x} \times \frac{d\pi x}{dx}$ $=-\sin{\sqrt{\sin(\tan{\pi x})}} \times \frac{1}{2\sqrt{\sin(\tan{\pi x})}} \times \cos(\tan{\pi x}) \times \sec^2{\pi x} \times \pi$ Simplifying the above expression, $y' =-\frac{1}{2}\pi\frac{\sin{\sqrt{\sin(\tan{\pi x})}} \times \cos(\tan{\pi x}) \times \sec^2{\pi x}}{\sqrt{\sin(\tan{\pi x})}}$