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: 14

Answer

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

Work Step by Step

\[\begin{gathered} y = {e^{\sqrt x }} \hfill \\ \hfill \\ use\,\,\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\,\,\, \hfill \\ \hfill \\ set\,\,u = \sqrt x \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {e^{\sqrt x }}\,{\left( {\sqrt x } \right)^,} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {e^{\sqrt x }}\,\left( {\frac{1}{{2\sqrt x }}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }} \hfill \\ \end{gathered} \]
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