University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 5



Work Step by Step

Given that $y=(x+1)^2(x^2+2x)$ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\frac{d}{dx}[(x+1)^2(x^2+2x)]$ $=2(x+1)(x^2+2x)+(x+1)^2(2x+2)$ $=(x+1)[2(x^2+2x)+(x+1)(2x+2)$ Hence, $y'=(2x+2)(2x^2+4x+1)$
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