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.