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 - Page 160: 30

Answer

$y'= 2s-2$

Work Step by Step

Quotient Rule: $y = \frac{4s^3-8s^2+4s}{4s}$ $y'=\frac{(12s^2-16s+4)(4s)-(4s^3-8s^2+4s)(4)}{(4s)^2}= 2s-2$ Division Then Power Rule: $y = \frac{4s^3-8s^2+4s}{4s}=s^2-2s+1$ $y'=1(2)s^{2-1}-2(1)s^{1-1}+1(0)s^{0-1} = 2s-2$
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