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