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: 9

Answer

$y' = 12x - 10 + 10x^{-3}$ $y'' = 12 -30x^{-4}$

Work Step by Step

$y = 6x^{2} - 10x - 5x^{-2}$ $y' = 12x - 10 + 10x^{-3}$ $y'' = 12 - 0 -30x^{-4}$
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