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Question
If \[y = \sqrt{\sin x + y}, \text { then }\frac{dy}{dx} \text { equals }\] ______________ .
Options
\[\frac{\cos x}{2y - 1}\]
\[\frac{\cos x}{1 - 2y}\]
\[\frac{\sin x}{1 - 2y}\]
\[\frac{\sin x}{2y - 1}\]
Solution
\[\frac{\cos x}{2y - 1}\]
\[\text { We have, y } = \sqrt{\sin x + y}\]
\[\text { Squaring both sides, we get }, \]
\[ y^2 = \sin x + y\]
\[ \Rightarrow y^2 - y = \sin x\]
\[ \Rightarrow 2y\frac{dy}{dx} - \frac{dy}{dx} = \cos x\]
\[ \Rightarrow \frac{dy}{dx}\left( 2y - 1 \right) = \cos x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos x}{2y - 1}\]
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