Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 42

Suppose that $g(\theta) = \theta \sin \theta$, find $g''\left(\frac{\pi}{6}\right)$


$\begin{equation} \begin{aligned} g'(\theta) &= \frac{d}{d \theta} ( \theta \sin \theta )\\ \\ g'(\theta) &= (\theta) \frac{d}{d\theta} (\sin \theta) + (\sin \theta) \frac{d}{d\theta}(\theta)\\ \\ g'(\theta) &= (\theta) (\cos \theta) + (\sin \theta) (1)\\ \\ g'(\theta) &= \theta \cos \theta + \sin \theta\\ \\ g''(\theta) &= \frac{d}{d \theta} ( \theta \cos \theta ) + \frac{d}{d \theta} (\sin \theta)\\ \\ g''(\theta) &= \left[ (\theta) \frac{d}{d\theta} (\cos \theta) + (\cos \theta) \frac{d}{d\theta}(\theta) \right] + \cos \theta\\ \\ g''(\theta) &= (\theta)(-\sin \theta) + (\cos \theta) (1) + \cos \theta\\ \\ g''(\theta) &= - \theta \sin \theta + \cos \theta + \cos \theta\\ \\ g''(\theta) &= - \theta \sin \theta + 2 \cos \theta\\ \\ g''\left(\frac{\pi}{6}\right) &= 2 \cos \theta - \theta \sin \theta\\ \\ g''\left(\frac{\pi}{6}\right) &= 2 \cos \frac{\pi}{6} - \left(\frac{\pi}{6}\right)\left(\sin\frac{\pi}{6}\right)\\ \\ g''\left(\frac{\pi}{6}\right) &= (\cancel{2}) \left(\frac{\sqrt{3}}{\cancel{2}}\right) - \left(\frac{\pi}{6}\right) \left(\frac{1}{2}\right)\\ \\ g''\left(\frac{\pi}{6}\right) &= \sqrt{3} - \frac{\pi}{12} \qquad \text{or} \qquad g''\left(\frac{\pi}{6}\right) = 1.47025142 \end{aligned} \end{equation}$