University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 198: 21



Work Step by Step

We evaluate the function: $y=\frac{2x}{{(1+x^2)}}$ on differentiating the above: $\frac{dy}{dx}=\frac{d\frac{2x}{(1+x^2)}}{dx}$ or ${dy}=(\frac{2(1+x^2)-2x\times2x}{(1+x^2)^2}){dx}$ or ${dy}=(\frac{(2+2x^2)-4x^2}{(1+x^2)^2}){dx}$ or ${dy}=(\frac{2-2x^2)}{(1+x^2)^2}){dx}$ The final answer is: ${dy}=(\frac{2-2x^2)}{(1+x^2)^2}){dx}$
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