Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises: 27

Answer

$h'(x)=4x^{3}+4x$

Work Step by Step

$h(x)=(x^{2}+1)^{2}$ $h(x)=x^{4}+2x^{2}+1$ $h'(x)=4x^{4-1}+2x^{2-1}$ $h'(x)=4x^{3}+4x^{1}$ $h'(x)=4x^{3}+4x$ OR $h(x)=(x^{2}+1)^{2}$ $h(x)=(x^{2}+1)(x^{2}+1)$ $u=x^{2}+1$ $u'=2x^{2-1}+0$ $u'=2x^{1}$ $u'=2x$ $v=u=x^{2}+1$ $v'=u'=2x$ $h'(x)=u'v+uv'$ $h'(x)=(2x)(x^{2}+1)+(x^{2}+1)(2x)$ $h'(x)=2x^{3}+2x+2x^{3}+2x$ $h'(x)=4x^{3}+4x$
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