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 - Page 200: 16

Answer

\[y' = \frac{{2\sqrt y }}{{1 - 4\sqrt y }}\]

Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\] \[1 + 2y' = \frac{1}{{2\sqrt y }} \cdot y'\] \[then\] \[2y' - \frac{{y'}}{{2\sqrt y }} = - 1\] \[factor\,\,out\,\,y'\] \[\begin{gathered} y'\,\left( {2 - \frac{1}{{2\sqrt y }}} \right) = - 1 \hfill \\ \hfill \\ y' = - \frac{1}{{2 - \frac{1}{{2\sqrt y }}}} \hfill \\ \hfill \\ \end{gathered} \] \[\begin{gathered} simplify \hfill \\ \hfill \\ y' = \frac{{2\sqrt y }}{{1 - 4\sqrt y }} \hfill \\ \end{gathered} \]
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