## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 39

#### Answer

$$f''(\theta) =2\cos \theta-\theta \sin \theta.$$

#### Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\sin x)'=\cos x$. Recall that $(\cos x)'=-\sin x$. Since $f(\theta)=\theta \sin \theta$, then using the product rule we have $$f'(\theta)= (\theta)' \sin \theta +\theta (\sin \theta )'\\ =\sin \theta + \theta \cos \theta,$$ and again using the product rule, we have $$f''(\theta)=\cos \theta+(\theta)' \cos \theta+\theta (\cos \theta)' \\= \cos \theta + \cos \theta-\theta \sin \theta\\ =2\cos \theta-\theta \sin \theta.$$

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