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