Question

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

• A. $\frac{\cos x}{1-2y}$
• B. $\frac{\cos x}{2y-1}$
• C. $\frac{\sin x}{1-2y}$
• D. $\frac{\sin x}{2y-1}$

Answer 1

The correct option is B $\frac{\cos x}{2y-1}$

Given that,
$y=\sqrt{\sin x+y}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\Rightarrow y^2=\sin x+y\mathrm{Differentiating}w.r.t.x\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\Rightarrow 2y\frac{\mathrm{dy}}{\mathrm{dx}}=\cos x+\frac{\mathrm{dy}}{\mathrm{dx}}\Rightarrow \left(2y-1\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\cos x\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos x}{\left(}$