## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 29

#### Answer

$$H'(\theta)=\sin\theta+\theta\cos\theta$$ $$H''(\theta)=2\cos\theta-\theta\sin\theta$$

#### Work Step by Step

$$H(\theta)=\theta\sin\theta$$ 1) Find $H'(\theta)$ $$H'(\theta)=(\theta\sin\theta)'$$ Apply the Product Rule: $$H'(\theta)=(\theta)'\sin\theta+\theta(\sin\theta)'$$ $$H'(\theta)=1\times\sin\theta+\theta\cos\theta$$ $$H'(\theta)=\sin\theta+\theta\cos\theta$$ 2) Find $H''(\theta)$ $$H''(\theta)=(\sin\theta+\theta\cos\theta)'$$ $$H''(\theta)=(\sin\theta)'+(\theta\cos\theta)'$$ Again, apply the Product Rule: $$H''(\theta)=(\cos\theta)+((\theta)'\cos\theta+\theta(\cos\theta)')$$ $$H''(\theta)=\cos\theta+\cos\theta-\theta\sin\theta$$ $$H''(\theta)=2\cos\theta-\theta\sin\theta$$

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