Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 26

Answer

$g'(w)=25r^{4}+54r^{2}+2r+9$

Work Step by Step

$g(w)=(5r^{3}+3r+1)(r^{2}+3)$ $u=5r^{3}+3r+1$ $u'=5(3)r^{3-1}+3r^{1-1}+0$ $u'=15r^{2}+3r^0$ $u'=15r^{2}+3$ $v=r^{2}+3$ $v'=2r^{2-1}+0$ $v'=2r^{1}$ $v'=2r$ $g'(w)=u'v+uv'$ $g'(w)=(15r^{2}+3)(r^{2}+3)+(5r^{3}+3r+1)(2r)$ $g'(w)=15r^{4}+45r^{2}+3r^{2}+9+10r^{4}+6r^{2}+2r$ $g'(w)=25r^{4}+54r^{2}+2r+9$
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