If $\sqrt{y-\sqrt{y-\sqrt{y\dots \dots \mathrm{\infty }}}}=\sqrt{x+\sqrt{x+\sqrt{x\dots \dots \mathrm{\infty }}}}\text{ then }\frac{dy}{dx}$ is equal to

$\mathrm{a}\sqrt{y-\sqrt{y-\sqrt{y\dots \dots \mathrm{\infty }}}}=\sqrt{x+\sqrt{x+\dots \mathrm{\infty }}}=v$

$\Rightarrow \sqrt{y-v}=\sqrt{x+v}=v$

$\left.\Rightarrow \begin{array}{l}y=v^2+v\\ x=v^2-v\end{array}\right\}\Rightarrow \frac{y+x}{2}=v^2\text{ and }\frac{y-x}{2}=v$

$\frac{dy}{dv}=2v+1,\frac{dx}{dv}=2v-1$

$\frac{dy}{dx}=\frac{dy/dv}{dx/dv}=\frac{2v+1}{2v-1}$

$\frac{dy}{dx}=\frac{y-x+1}{y-x-1}$