Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 191: 18

Answer

\[\frac{{dy}}{{dx}} = - 2x{e^{ - {x^2}}}\]

Work Step by Step

\[\begin{gathered} y = {e^{ - {x^2}}} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set\,\,u = - {x^2} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {e^{ - {x^2}}}\frac{d}{{dx}}\,\left( { - {x^2}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = - 2x{e^{ - {x^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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.