Question

If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+..\mathrm{to}\infty }}}$, prove that $\left(2y-1\right)\frac{dy}{dx}=\frac{1}{x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+...\mathrm{to}\infty }}} \\ \Rightarrow y=\sqrt{\log x+y} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ {y}^2=\log x+y \\ \Rightarrow 2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(2y-1\right)=\frac{1}{x} \end{gathered}$$