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