University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 17

Answer

$y' = \frac{-19}{(3x-2)^{2}}$

Work Step by Step

$y = \frac{2x+5}{3x-2}$ Using quotient rule: $y' = \frac{(3x-2)(2) - (2x+5)(3)}{(3x-2)^{2}}$ $y' = \frac{-19}{(3x-2)^{2}}$
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