Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 18

Answer

$$dy = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx$$

Work Step by Step

$$\eqalign{ & y = \frac{{2x}}{{1 + {x^2}}} \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{2x}}{{1 + {x^2}}}} \right] \cr & {\text{Use the quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + {x^2}} \right)\left( 2 \right) - 2x\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{2 + 2{x^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{Simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{Write in differential form}} \cr & dy = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx \cr} $$
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.