Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.3 Exercises - Page 125: 32



Work Step by Step

$h(x)=(x^2+3)^3=(x^2+3)^2(x^2+3)=(x^4+6x^2+9)(x^2+3)$ Product Rule $h'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=x^4+6x^2+9 ;u’(x)=4x^3+12x $ $v(x)=x^2+3 ;v’(x)= 2x$ $h'(x)=(4x^3+12x)(x^2+3)+(2x)(x^4+6x^2+9)$ $=2x^5+12x^3+18x+4x^5+12x^3+12x+36x$ $=6x^5+36x^3+54x=6x(x^4+6x^2+9)$ $=(6x)(x^2+3)^2$
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