University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 71



Work Step by Step

Simplify: $\dfrac{d(\dfrac{(7x-2)^4}{28}+C)}{dx}$ Thus, we have $\dfrac{d(\dfrac{(7x-2)^4}{28}+C)}{dx}=(4)[\dfrac{(7x-2)^3}{28}](7)=(7x-2)^3$
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