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: 2


The derivative is $12x^3-12x^2+15$.

Work Step by Step

Product Rule: y'=$((u(x)(v(x))’=u'(x)v(x)+u(x)v’(x))$ $u(x)=(3x-4) ;u’(x)=(3) $ $v(x)=(x^3+5) ;v’(x)=(3x^2) $ $y'=(3)(x^3+5)+(3x-4)(3x^2)=$ $(3x^3+15)+(9x^3-12x^2)=$ $12x^3-12x^2+15$
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