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 - Page 136: 70

Answer

$f'(x)=-\dfrac{13}{(2x-5)^2}, f'(9)=-\dfrac{1}{13}.$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x+4; u'(x)=1$ $v(x)=2x-5; v'(x)=2$ $f'(x)=\dfrac{1(2x-5)-2(x+4)}{(2x-5)^2}=-\dfrac{13}{(2x-5)^2}.$ $f'(9)=-\dfrac{13}{(2(9)-5)^2}=-\dfrac{1}{13}.$ A graphing utility was used to verify this result.
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