University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 35


a) $y=\dfrac{1}{x}+C$ b) $y=x+\dfrac{1}{x}+C$ c) $y=5x-\dfrac{1}{x}+C$

Work Step by Step

a) When $(\dfrac{1}{x})'=-\dfrac{1}{x^2}$ Thus, $y'=-\dfrac{1}{x^2} \implies y=\dfrac{1}{x}+C$ b) When $(x+\dfrac{1}{x})'=1-\dfrac{1}{x^2}$ Thus, $y'=1-\dfrac{1}{x^2} \implies y=x+\dfrac{1}{x}+C$ c) When $(5x-\dfrac{1}{x})'=5+\dfrac{1}{x^2}$ Thus, $y'=5+\dfrac{1}{x^2} \implies y=5x-\dfrac{1}{x}+C$
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