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: 18

Answer

$h'(z)=4z^3+9z^2-6z-1$

Work Step by Step

Product Rule: $h(z) = (z^3+4z^2+z)(z-1)$ $h'(z) = (3z^2+8z+1)(z-1) + (z^3+4z^2+z)(1) =4z^3+9z^2-6z-1$ Expansion Then Power Rule: $h(z) = (z^3+4z^2+z)(z-1) = (z^4+3z^3-3z^2-z)$ $h'(z) = (4)z^{4-1}+3(3)z^{3-1}-3(2)z^{2-1}-(1)z^{1-1} = 4z^3+9z^2-6z-1$
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