Answer
$f'(x)=18x^{2}+6x+4$
Work Step by Step
$f(x)=(2x+1)(3x^{2}+2)$
$u=2x+1$
$u'=2$
$v=3x^{2}+2$
$v'=3(2)x^{2-1}+0$
$v'=6x$
$f(x)=u'v+uv'$
$f'(x)=2(3x^{2}+2)+(2x+1)(6x)$
$f'(x)=6x^{2}+4+12x^{2}+6x$
$f'(x)=18x^{2}+6x+4$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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: 25
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