If
\(\sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \ldots \infty}}}\) \(=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots \infty}}}\) then \(\frac{d y}{d x}=\)

\(\text { Let } \sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \infty}}}=\) \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \infty}}}=\mathrm{z} \)
\( \Rightarrow \mathrm{y}-\mathrm{z}=\mathrm{z}^2 \text { and } \mathrm{x}+\mathrm{z}=\mathrm{z}^2 \)
\( \Rightarrow \mathrm{y}=\mathrm{z}^2+\mathrm{z} \text { and } \mathrm{x}=\mathrm{z}^2-\mathrm{z} \)
\( \Rightarrow \frac{\mathrm{dy}}{\mathrm{dz}}=2 \mathrm{z}+1 \text { and } \frac{\mathrm{dx}}{\mathrm{dz}}=2 \mathrm{z}-1 \text { also } 2 \mathrm{z}=\mathrm{y}-\mathrm{x} \)
\( \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{z}+1}{2 \mathrm{z}-1}=\frac{\mathrm{y}-\mathrm{x}+1}{\mathrm{y}-\mathrm{x}-1}\)