Answer
\[\frac{{dy}}{{dx}} = - \frac{{\cos x}}{{\,{{\left( {2 + \sin x} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{1}{{2 + \sin x}} \hfill \\
\hfill \\
by\,\,the\,\,quotient\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\left( {2 + \sin x} \right)\left[ 1 \right]' - \left( 1 \right)\left[ {2 + \sin x} \right]'}}{{{{\left( {2 + \sin x} \right)}^2}}} \hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{0 - \,\left( {\cos x} \right)}}{{\,{{\left( {2 + \sin x} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = - \frac{{\cos x}}{{\,{{\left( {2 + \sin x} \right)}^2}}} \hfill \\
\end{gathered} \]
