Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 148: 35



Work Step by Step

Apply the Chain Rule, we get $f'(\theta)=2(\dfrac{\sin\theta}{1+\cos\theta})(\dfrac{\sin\theta}{1+\cos\theta})'$ ...(1) and $(\dfrac{\sin\theta}{1+\cos\theta})'=\dfrac{(\sin\theta)'(1+\cos\theta)-\sin\theta(1+\cos\theta)'}{(1+\cos\theta)^2}$ or, $\dfrac{\cos\theta(1+\cos\theta)-\sin\theta(-\sin\theta)}{(1+\cos\theta)^2}=\dfrac{1}{1+\cos\theta}$ Now, equation (1) becomes: $f'(\theta)=2(\dfrac{\sin\theta}{1+\cos\theta})[\dfrac{1}{1+\cos\theta}]=\dfrac{2\sin\theta}{(1+\cos\theta)^2}$
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