Calculus 10th Edition

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

Chapter 2 - Differentiation - Review Exercises: 34

Answer

$f'(x)=-\dfrac{2(x^2+7x-4)}{(x^2+4)^2}.$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=2x+7; u'(x)=2$ $v(x)=x^2+4; v'(x)=2x$ $f'(x)=\dfrac{2(x^2+4)-2x(2x+7)}{(x^2+4)^2}$ $=-\dfrac{2(x^2+7x-4)}{(x^2+4)^2}.$
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