Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 10

Differentiate $\displaystyle \frac{1 + \sin x}{x + \cos x}$


$\begin{equation} \begin{aligned} y' =& \frac{\displaystyle (x + \cos x) \frac{d}{dx} (1 + \sin x) - \left[ (1 + \sin x) \frac{d}{dx} (x + \cos x) \right] }{(x + \cos x)^2} && \text{Using Quotient Rule} \\ \\ y' =& \frac{(x + \cos x)(0 + \cos x) - [(1 + \sin x) (1 - \sin x)]}{(x + \cos x)^2} && \text{Simplify the equation} \\ \\ y' =& \frac{x \cos x + \cos^2 x - (1 - \sin^2 x)}{(x + \cos x)^2} && \text{} \\ \\ y' =& \frac{x \cos x + \cos^2 x - 1 + \sin^2 x}{(x + \cos x)^2} && \text{Use the Trigonometric Identities to simplify the equation} \\ \\ y' =& \frac{x \cos x -1 + 1}{(x + \cos x)^2} && \text{Combine like terms} \\ \\ y' =& \frac{x \cos x}{(x + \cos x)^2} \end{aligned} \end{equation}$