Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5: 23

Answer

The Derivative is: $\frac{dr}{d\theta}=-\theta(2\sin\theta+\theta\cos\theta)$

Work Step by Step

$r=4-\theta^2\sin\theta$ Applying Derivative rules: $y'=f'(x)+g'(x)$ $and$ $g'(x)=h'(x)\cdot v(x)+h(x)\cdot v'(x)$ $\frac{dr}{d\theta}=\frac{d}{d\theta}(4)-\frac{d}{d\theta}(\theta^2\sin\theta)$ $\frac{dr}{d\theta}=(0)-((2\theta^{2-1})(\sin\theta)+\theta^2(\cos\theta)$ $\frac{dr}{d\theta}=-2\theta\sin\theta-\theta^2\cos\theta$ $\frac{dr}{d\theta}=-\theta(2\sin\theta+\theta\cos\theta)$
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