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: 24

Answer

\[y' = - \frac{1}{{2y - 2\cos y\sqrt {x + {y^2}} }}\]

Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\] \[\frac{1}{{2\sqrt {x + {y^2}} }}\, \cdot \,\,\,\,{\left( {x + {y^2}} \right)^\prime } = \cos y \cdot y'\] \[then\] \[\frac{1}{{2\sqrt {x + {y^2}} \, \cdot \,\left( {1 + 2yy'} \right)}} = y'\cos y\] \[simplify\] \[\frac{{2yy'}}{{2\sqrt {x + {y^2}} }} - y'\cos y = - \frac{1}{{2\sqrt {x + {y^2}} }}\] \[factor\,\,y'\] \[y'\,\left( {\frac{y}{{\sqrt {x + {y^2}} }} - \cos y} \right) = - \frac{1}{{2\sqrt {x + {y^2}} }}\] \[solve\,\,for\,\,y'\] \[\begin{gathered} y' = - \frac{{\frac{1}{{2\sqrt {x + {y^2}} }}}}{{\frac{y}{{\sqrt {x + {y^2}} }} - \cos y}} \hfill \\ \hfill \\ y' = - \frac{1}{{2y - 2\cos y\sqrt {x + {y^2}} }} \hfill \\ \end{gathered} \]
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.