Question:
Choose the correct alternative in the following:
If $y=\sqrt{\sin x+y}$, then $\frac{d y}{d x}=$
A. $\frac{\sin x}{2 y-1}$
B. $\frac{\sin x}{1-2 y}$
C. $\frac{\cos x}{1-2 y}$
D. $\frac{\cos x}{2 y-1}$
Solution:
$y=\sqrt{\sin x+y}$
Squaring both sides
$\Rightarrow y^{2}=\sin x+y$
Differentiating w.r.t $x$ we get,
$\Rightarrow 2 y \cdot \frac{d y}{d x}=\cos x+\frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}(2 y-1)=\cos x$
$\Rightarrow \frac{d y}{d x}=\frac{\cos x}{2 y-1}=D$
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