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: 14

Answer

$y' = 30x^{2} +14x -12$

Work Step by Step

Using product rule: $y = (2x+3)(5x^{2} -4x)$ $y' = (2x+3)(10x-4) + (5x^{2} -4x)(2)$ $y' = 30x^{2} +14x -12$ Using sum of simpler terms: $y = 10x^{3} + 7x^{2} -12x$ $y' = 30x^{2} +14x -12$
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