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 199: 33



Work Step by Step

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