#### Answer

\[\frac{{dy}}{{dx}} = \frac{1}{{\sin 2x - 1}}\]

#### Work Step by Step

\[\begin{gathered}
y = \frac{{\sin x}}{{\sin x - \cos x}} \hfill \\
\hfill \\
using\,\,quotient\,\,\,rule\,\,for\,\,derivatives \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\cos x\,\left( {\sin x - \cos x} \right) - \sin x\,\left( {\cos x + \sin x} \right)}}{{\,{{\left( {\sin x - \cos x} \right)}^2}}} \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\cos x\sin x - {{\cos }^2}x - \sin x\cos x - {{\sin }^2}x}}{{{{\sin }^2}x - 2\sin x\cos x + {{\cos }^2}x}} \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
use\,\,the\,\,identity\,\,:\,\,{\cos ^2}x + {\sin ^2}x = 1 \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{ - 1}}{{1 - 2\sin x\cos x}} = \frac{1}{{2\sin x\cos x - 1}} \hfill \\
\hfill \\
using\,:\,\,\sin 2x = 2\sin x\cos x \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{1}{{\sin 2x - 1}} \hfill \\
\hfill \\
\end{gathered} \]