Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 14

Differentiate $\displaystyle y = \csc \theta (\theta + \cot \theta)$


$\begin{equation} \begin{aligned} y' =& (\csc \theta) \frac{d}{d\theta} (\theta + \cot \theta) + (\theta + \cot \theta) \frac{d}{d\theta} (\csc \theta) && \text{Using Product Rule} \\ \\ y' =& (\csc \theta)(1 - \csc^2 \theta) + (\theta + \cot \theta)(- \csc \theta \cot \theta) && \text{Using Trigonometric Identities} \\ \\ y' =& (\csc \theta) (- \cot ^ 2 \theta) + (\theta + \cot \theta) (- \csc \theta \cot \theta) && \text{Simplify the equation} \\ \\ y' =& - \csc \theta \cot^2 \theta - \theta \csc \theta \cot \theta - \csc \theta \cot^2 \theta && \text{Combine like terms} \\ \\ y' =& -2 \csc \theta \cot^2 \theta - \theta \csc \theta \cot \theta \\ \\ & \text{or} \\ \\ y' =& - \csc \theta \cot \theta (2 \cot \theta + \theta) && \text{} \end{aligned} \end{equation}$