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