Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - Review - Exercises - Page 197: 42


$$g''(\pi/6) =\sqrt{3}-\frac{\pi }{12} $$

Work Step by Step

Given $$g(\theta)=\theta\sin \theta$$ Since \begin{align*} g'(\theta)&=\theta\cos \theta+ \sin \theta\\ g''(\theta)&=-\theta\sin \theta+ 2\cos \theta \end{align*} then $$g''(\pi/6) =-(\pi/6)\sin(\pi/6)+ 2\cos (\pi/6)=\sqrt{3}-\frac{\pi }{12} $$
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