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: 25

Answer

$f'(x)=18x^{2}+6x+4$

Work Step by Step

$f(x)=(2x+1)(3x^{2}+2)$ $u=2x+1$ $u'=2$ $v=3x^{2}+2$ $v'=3(2)x^{2-1}+0$ $v'=6x$ $f(x)=u'v+uv'$ $f'(x)=2(3x^{2}+2)+(2x+1)(6x)$ $f'(x)=6x^{2}+4+12x^{2}+6x$ $f'(x)=18x^{2}+6x+4$
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