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: 13

Answer

$f'(x)=15x^4+8x^3+21x^2+16x-20$ $f'(0)=-20$

Work Step by Step

Product Rule $f'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=x^3+4x ;u’(x)=3x^2+4 $ $v(x)=(3x^2+2x-5) ;v’(x)=(6x+2) $ $f'(x)=(3x^2+4)(3x^2+2x-5)+(x^3+4x)(6x+2)=$ $(9x^4+6x^3-15x^2+12x^2+8x-20)+(6x^4+24x^2+2x^3+8x)=$ $15x^4+8x^3+21x^2+16x-20$ $f'(0)=15(0)^4+8(0)^3+21(0)^2+16(0)-20=-20$
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