Solve this


If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \text { to } \infty}}}$, prove that $(2 y-1) \frac{d y}{d x}=\frac{1}{x}$


$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\cdots \text { to } \infty}}}$

$y=\sqrt{\log x+y}$

On squaring both sides,

$y^{2}=\log x+y$

Differentiating both sides with respect to $x$,

$2 y \frac{d y}{d x}=\frac{1}{x}+\frac{d y}{d x}$

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

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

Hence proved.

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