Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - Review Exercises: 37

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} $ $y'=(6x)(\sec{x})+(\sec{x}\tan{x})(3x^2)$ $=3x(2+x\tan{x})\sec{x}.$
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