Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 15

Answer

$$f'(\theta)=\frac{1}{2}\sin{2\theta}+\theta\cos{2\theta}$$

Work Step by Step

$$f(\theta)=\theta\cos\theta\sin\theta$$ $$f(\theta)=(\theta\cos\theta)\sin\theta$$ Apply the Product Rule with 2 elements $\theta\cos\theta$ and $\sin\theta$ first: $$f'(\theta)=(\theta\cos\theta)'\sin\theta+(\theta\cos\theta)(\sin\theta)'$$ $$f'(\theta)=(\theta\cos\theta)'\sin\theta+(\theta\cos\theta)\cos\theta$$ $$f'(\theta)=(\theta\cos\theta)'\sin\theta+\theta\cos^2\theta$$ Now apply the Product Rule for $(\theta\cos\theta)'$: $$f'(\theta)=[(\theta)'\cos\theta+\theta(\cos\theta)']\sin\theta+\theta\cos^2\theta$$ $$f'(\theta)=[\cos\theta-\theta\sin\theta]\sin\theta+\theta\cos^2\theta$$ $$f'(\theta)=\sin\theta\cos\theta-\theta\sin^2\theta+\theta\cos^2\theta$$ $$f'(\theta)=\frac{1}{2}\sin{2\theta}+\theta(\cos^2\theta-\sin^2\theta)$$ $$f'(\theta)=\frac{1}{2}\sin{2\theta}+\theta\cos{2\theta}$$

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