Question

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

(a) $\frac{\sin x}{2y-1}$

(b) $\frac{\sin x}{1-2y}$

(c) $\frac{\cos x}{1-2y}$

(d) $\frac{\cos x}{2y-1}$

Answer 1

(d) $\frac{\cos x}{2y-1}$

$$\begin{gathered} \mathrm{We}\mathrm{have},y=\sqrt{\sin x+y} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ {y}^2=\sin x+y \\ \Rightarrow y^2-y=\sin x \\ \Rightarrow 2y\frac{dy}{dx}-\frac{dy}{dx}=\cos x \\ \Rightarrow \frac{dy}{dx}\left(2y-1\right)=\cos x \\ \Rightarrow \frac{dy}{dx}=\frac{\cos x}{2y-1} \end{gathered}$$