Question

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

Answer 1

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