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: 15

Answer

\[y' = - \frac{1}{{1 + \sin y}}\]

Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\] \[1 + y' = - \sin y \cdot y'\] \[then\] \[y' + y'\sin y = - 1\] \[factor\] \[\begin{gathered} y'\,\left( {1 + \sin y} \right) = - 1 \hfill \\ \hfill \\ y' = - \frac{1}{{1 + \sin y}} \hfill \\ \hfill \\ \end{gathered} \]
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