If \[y=\sqrt{\sin x+y},\] then \[\frac{dy}{dx}\] equals to [RPET 2001]

1.	If \[y=\sqrt{\sin x+y},\] then \[\frac{dy}{dx}\] equals to [RPET 2001]
A.	\[\frac{\sin x}{2y-1}\]
B.	\[\frac{\cos x}{2y-1}\]
C.	\[\frac{\sin x}{2y+1}\]
D.	\[\frac{\cos x}{2y+1}\]
Answer» C. \[\frac{\sin x}{2y+1}\]

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If \[y=\sqrt{\sin x+y},\] then \[\frac{dy}{dx}\] equals to [RPET 2001]

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