Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises: 14

Answer

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

Work Step by Step

\[\begin{gathered} {e^{xy}} = 2y \hfill \\ \hfill \\ use\,\,the\,\,implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {{e^{xy}}} \right] = \frac{d}{{dx}}\,\,\left[ {2y} \right] \hfill \\ \end{gathered} \] \[differentiate\] \[{e^{xy}}\,\,\left[ {xy' + y} \right] = 2y'\] \[multiply\] \[xy'\,{e^{xy}} + y{e^{xy}} = 2y'\] \[collect\,\,like\,\,terms\] \[xy'{e^{xy}} - 2y' = - y{e^{xy}}\] \[solve\,\,for\,\,{y^,}\] \[\begin{gathered} y'\,\left( {x{e^{xy}} - 2} \right) = - y{e^{xy}} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{y{e^{xy}}}}{{2 - x{e^{xy}}}} \hfill \\ \end{gathered} \]
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