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: 27

Answer

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

Work Step by Step

$y = \frac{1}{(x^{2}-1)(x^{2}+x+1)}$ $y = \frac{1}{x^{4} + x^{3} - x-1}$ $y' = \frac{0-(4x^{3}+3x^{2}-1)}{(x^{4}+x^{3}-x-1)^{2}}$ $y' = \frac{-4x^{3}-3x^{2}+1}{(x^{4}+x^{3}-x-1)^{2}}$ $y' = \frac{-4x^{3}-3x^{2}+1}{((x^2-1)(x^2+x+1))^2}$
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