If $y = \sqrt {\sin x + y}$, then $\frac{{dy}}{{dx}}$ is equal to

therefore,$\quad \frac{{dy}}{{dx}} = \frac{1}{2}{(\sin x + y)^{ - 1/2}} \cdot \frac{d}{{dx}}(\sin x + y)$

$\Rightarrow$ $\quad \frac{{dy}}{{dx}} = \frac{1}{2} \cdot \frac{1}{{{{(\sin x + y)}^{1/2}}}} \cdot \left( {\cos x + \frac{{dy}}{{dx}}} \right)$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{1}{{2y}}\left( {\cos x + \frac{{dy}}{{dx}}} \right)$

$\Rightarrow$ $\frac{{dy}}{{dx}}\left( {1 - \frac{1}{{2y}}} \right) = \frac{{\cos x}}{{2y}}$

therefore,$\frac{{dy}}{{dx}} = \frac{{\cos x}}{{2y}} \cdot \frac{{2y}}{{2y - 1}} = \frac{{\cos x}}{{2y - 1}}$