Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 140: 31



Work Step by Step

$y=\frac{x^2+1}{x^3-1}$ Use the quotient rule. $y'=\frac{(x^3-1)(2x)-(x^2+1)(3x^2)}{(x^3-1)^2}$ Distribute. $y'=\frac{2x^4-2x-3x^4-3x^2}{(x^3-1)^2}$ Combine like terms. $y'=\frac{-x^4-3x^2-2x}{(x^3-1)^2}$ Factor out the x. $\frac{x(-x^3-3x-2)}{(x^3-1)^2}$
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