#### Answer

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

#### Work Step by Step

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