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: 34



Work Step by Step

$g(x)=x^2(\frac{2(x+1)-x}{x(x+1)})=\frac{x^3+2x^2}{x^2+x}$ Using the quotient rule: $gā€™(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^3+2x^2; u'(x)=3x^2+4x$ $v(x)=x^2+x; v'(x)=2x+1$ $g'(x)\frac{(3x^2+4x)(x^2+x)-(2x+1)(x^3+2x^2)}{(x(x+1))^2}=$ $\frac{(3x^4+7x^3+4x^2)-(2x^4+5x^3+2x^2)}{x^2(x+1)^2=}$ $\frac{x^2(x^2+2x+2)}{x^2(x+1)^2}=\frac{x^2+2x+2}{(x+1)^2}$
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