University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 28

Answer

$y = \frac{-6(x^{2}-2)}{(x^{2}-3x+2)^{2}}$ $y = \frac{-6(x^{2}-2)}{((x-1)(x-2))^2}$

Work Step by Step

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