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: 22

Answer

\[y' = \frac{{\cos x\,\left( {1 - \cos y} \right)}}{{\sin y\,\left( {1 - \sin x} \right)}}\]

Work Step by Step

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