A) \[\frac{\cos x}{2y-1}\]
B) \[\frac{-\cos x}{2y-1}\]
C) \[\frac{\sin x}{1-2y}\]
D) \[\frac{-\sin x}{1-2y}\]
E) \[\frac{2\cos x}{2y-1}\]
Correct Answer: A
Solution :
\[y=\sqrt{\sin x+\sqrt{\sin x+....\infty }}\] \[\Rightarrow \] \[y=\sqrt{\sin x+y}\] \[\Rightarrow \] \[{{y}^{2}}=\sin x+y\] On differentiating w.r.t.\[x,\]we get \[2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\cos x}{2y-1}\]
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