Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 188: 13



Work Step by Step

$y=\dfrac{x^{2}+1}{x^{3}-1}$ Differentiate using the quotient rule: $y'=\dfrac{(x^{3}-1)(x^{2}+1)'-(x^{2}+1)(x^{3}-1)'}{(x^{3}-1)^{2}}=...$ $...=\dfrac{(x^{3}-1)(2x)-(x^{2}+1)(3x^{2})}{(x^{3}-1)^{2}}=...$ Evaluate the products and simplify $...=\dfrac{2x^{4}-2x-3x^{4}-3x^{2}}{(x^{3}-1)^{2}}=\dfrac{-x^{4}-3x^{2}-2x}{(x^{3}-1)^{2}}=-\dfrac{x^{4}+3x^{2}+2x}{(x^{3}-1)^{2}}$
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