Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 16

Answer

$y' = 9t^2+34t-28$

Work Step by Step

Product Rule: $y=(t^2+7t)(3t-4)$ $y'=(2t+7)(3t-4)+(t^2+7t)(3) = 9t^2 + 34t-28$ Expansion then Power Rule: $y=(t^2+7t)(3t-4) = 3t^3-4t^2+21t^2-28t = 3t^3+17t^2-28t$ $y'=3(3)t^{3-1}+17(2)t^{2-1}-28(1)t^{1-1}=9t^2 + 34t-28$
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