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