Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3 - Page 125: 27


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

Work Step by Step

$y=\frac{1}{(x^2-1)(x^2+x+1)}$ Using Quotient Rule to find the Derivative: $y'=\frac{f′(x)⋅g(x)−f(x)⋅g′(x)}{g^2(x)}$ $y'=\frac{(0)(x^2-1)(x^2+x+1)-(1)[((2)x^{2-1}-0)(x^2+x+1)+(x^2-1)((2)x^{2-1}+x^{1-1}+0)]}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-[2x(x^2+x+1)+(2x+1)(x^2-1)]}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-2x^3-2x^2-2x-2x^3+2x-x^2+1}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-4x^3-3x^2+1}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-4x^3-3x^2+1}{(x^2-1)^2(x^2+x+1)^2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.