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