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: 29

Answer

$g'(x)=\dfrac{(x+5)(4-2x^2-20x)}{(x^2+2)^3}$

Work Step by Step

Using the quotient rule: $gā€™(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(x+5)^2; u'(x)=u'(t)\times t'=2(x+5)$ $v(x)=(x^2+2)^2; v'(x)=v'(z)\times z'=4x(x^2+2)$ $g'(x)=\dfrac{(2(x+5))(x^2+2)^2-(4x(x^2+2))(x+5)^2}{(x^2+2)^4}$ $=\dfrac{(x^2+2)(x+5)(2(x^2+2)-4x(x+5))}{(x^2+2)^4}$ $=\dfrac{(x+5)(4-2x^2-20x)}{(x^2+2)^3}$
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