Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 259: 22

Answer

(a) $dy = \frac{-2}{(x-1)^2}~dx$ (b) $dy = -0.1$

Work Step by Step

(a) $y = \frac{x+1}{x-1}$ $\frac{dy}{dx} = \frac{(1)(x-1)-(x+1)(1)}{(x-1)^2}$ $\frac{dy}{dx} = \frac{x-1-x-1}{(x-1)^2}$ $\frac{dy}{dx} = \frac{-2}{(x-1)^2}$ $dy = \frac{-2}{(x-1)^2}~dx$ (b) $dy = \frac{-2}{(x-1)^2}~dx$ $dy = \frac{-2}{(2-1)^2}~(0.05)$ $dy = \frac{-2}{1}~(0.05)$ $dy = -0.1$
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