Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises - Page 65: 111

Answer

$\frac{2x(1-3x^2)}{(x^2+1)^5}$

Work Step by Step

Start with cancelling the common factor $(x^2+1)^3$, we have: $\frac{(x^2+1)^4(2x)-x^2(4)(x^2+1)^3(2x)}{(x^2+1)^8}=\frac{(x^2+1)(2x)-x^2(4)(2x)}{(x^2+1)^5}=\frac{2x(1-3x^2)}{(x^2+1)^5}$
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