Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 50

Answer

$$dy = \frac{1}{{x - 1}}dx$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \ln \left( {1 - x} \right) \cr & {\text{Differentiate }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {1 - x} \right)} \right] \cr & f'\left( x \right) = \frac{{ - 1}}{{1 - x}} \cr & f'\left( x \right) = \frac{1}{{x - 1}} \cr & {\text{Write in the form }}dy = f'\left( x \right)dx \cr & dy = \frac{1}{{x - 1}}dx \cr} $$
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