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: 26

Answer

$f'(x)=-\frac{5}{(x-2)^2}$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(x^2+5x+6); u'(x)=(2x+5)$ $v(x)=(x^2-4); v'(x)=2x$ $f'(x)=\frac{(2x+5)(x^2-4)-(2x)(x^2+5x+6)}{(x^2-4)^2}=$ $\frac{(2x^3+5x^2-8x-20)+(-2x^3-10x^2-20)}{(x-2)^2(x+2)^2}=$ $\frac{-5x^2-20x-20}{(x-2)^2(x+2)^2}=-\frac{5(x+2)^2}{(x+2)^2(x-2)^2}=-\frac{5}{(x-2)^2}$
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