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Question:

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

Solution:

Here,

$y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots} \text { to } \infty}}$

$y=\sqrt{x+y}$

On squaring both sides,

$y^{2}=x+y$

Differentiating both sides with respect to $x$,

$2 y \frac{d y}{d x}=1+\frac{d y}{d x}$

$\frac{d y}{d x}(2 y-1)=1$

$\frac{d y}{d x}=\frac{1}{2 y-1}$

Hence proved.

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