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: 14

Answer

$y'=5x^4-12x^3+6x^2+2x-3$ $ y'(2)=9$.

Work Step by Step

Product Rule $y'=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=(x^2-3x+2) ;u’(x)=(2x-3) $ $v(x)=(x^3+1) ;v’(x)=(3x^2) $ $y'=(2x-3)(x^3+1)+(x^2-3x+2)(3x^2)$ $=(2x^4+2x-3x^3-3)+(3x^4-9x^3+6x^2)$ $=5x^4-12x^3+6x^2+2x-3$ $y'(2)=5(2)^4-12(2)^3+6(2)^2+2(2)-3=9$
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