Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 32

Answer

$g'(x)=\dfrac{6(3x^2-2)^2(3x^2+9x+2)}{(2x+3)^4}$

Work Step by Step

Using the quotient rule: $gā€™(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(3x^2-2)^3; u'(x)=(3)(6x)(3x^2-2)^{3-1}=18x(3x^2-2)^2.$ $v(x)=(2x+3)^3; v'(x)=(3)(2)(2x+3)^{3-1}=6(2x+3)^2.$ $g'(x)=\dfrac{(18x(3x^2-2)^2)((2x+3)^3)-(6(2x+3)^2)((3x^2-2)^3)}{((2x+3)^3)^2}$ $=\dfrac{6(3x^2-2)^2(3x^2+9x+2)}{(2x+3)^4}$
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