Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 28

Determine the derivative of the function $\displaystyle y = \frac{\cos \pi x}{\sin \pi x + \cos \pi x}$



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