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If
$\sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \ldots \infty}}}=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots \infty}}}$ then $\frac{d y}{d x}=$
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$\sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \ldots \infty}}}=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots \infty}}}$ then $\frac{d y}{d x}=$
Solution:
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Verified Answer
The correct answer is:
$\frac{y-x+1}{y-x-1}$
$\begin{aligned} & \text { Let } \sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \infty}}}=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \infty}}}=\mathrm{z} \\ & \Rightarrow \mathrm{y}-\mathrm{z}=\mathrm{z}^2 \text { and } \mathrm{x}+\mathrm{z}=\mathrm{z}^2 \\ & \Rightarrow \mathrm{y}=\mathrm{z}^2+\mathrm{z} \text { and } \mathrm{x}=\mathrm{z}^2-\mathrm{z} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dz}}=2 \mathrm{z}+1 \text { and } \frac{\mathrm{dx}}{\mathrm{dz}}=2 \mathrm{z}-1 \text { also } 2 \mathrm{z}=\mathrm{y}-\mathrm{x} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{z}+1}{2 \mathrm{z}-1}=\frac{\mathrm{y}-\mathrm{x}+1}{\mathrm{y}-\mathrm{x}-1}\end{aligned}$
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