Answer
$y'=3x(2+x\tan{x})\sec{x}$
Work Step by Step
Product Rule $(y'=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=3x^2 ;u’(x)=6x $
$v(x)=\sec{x} ;v’(x)=\sec{x}\tan{x} $
$y'=(6x)(\sec{x})+(\sec{x}\tan{x})(3x^2)$
$=3x(2+x\tan{x})\sec{x}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - Review Exercises - Page 157: 37
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