Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises: 12

Answer

a. $dy = \frac{-du}{(1+3u)^2}$ b. $dy = 2\theta(sin 2\theta + \theta cos(2\theta)) d\theta$

Work Step by Step

a. $y = \frac{1+2u}{1+3u}$ $f'(u) = \frac{2(1+3u) - 3(1+2u)}{(1+3u)^2}$ $f'(u) = \frac{2+6u - 3-6u}{(1+3u)^2}$ $f'(u) = \frac{-1}{(1+3u)^2}$ $dy = f'(u)du$ $dy = \frac{-1}{(1+3u)^2}du$ $dy = \frac{-du}{(1+3u)^2}$ b. $y = \theta^2 sin 2\theta$ $f'(\theta) = 2\theta sin2\theta + 2\theta^2 cos(2\theta)$ $f'(\theta) = 2\theta(sin 2\theta + \theta cos(2\theta))$ $dy = f'(\theta)d\theta$ $dy = 2\theta(sin 2\theta + \theta cos(2\theta)) d\theta$
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