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: 44

Answer

\[\frac{{dy}}{{dx}} = - 4{e^x}\,{\left( {1 - {e^x}} \right)^3}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {1 - {e^x}} \right)^4} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ {\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set\,\,u = 1 - {e^x} \hfill \\ \hfill \\ \frac{{du}}{{dx}} = - {e^x} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 4\,{\left( {1 - {e^x}} \right)^3}\,\left( { - {e^x}} \right) \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = - 4{e^x}\,{\left( {1 - {e^x}} \right)^3} \hfill \\ \end{gathered} \]
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