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: 15

Answer

$f'(x)=\frac{x^2-6x+4}{(x-3)^2}; f'(1)=-\frac{1}{4}$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(x^2-4); u'(x)=(2x)$ $v(x)=(x-3); v'(x)=(1)$ $f'(x)=\frac{(2x)(x-3)-(x^2-4)(1)}{(x-3)^2}=\frac{x^2-6x+4}{(x-3)^2}$ $f'(1)=\frac{(1)^2-6(1)+4}{((1)-3)^2}=-\frac{1}{4}$
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