If $\mathrm{y}=\sqrt{x+\sqrt{\mathrm{y}+\sqrt{x+\sqrt{\mathrm{y}+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \infty}}}}$, then $\frac{\mathrm{dy}}{\mathrm{d} x}=$

The infinite nested radical $y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \cdots}}}}$ contains a repeating pattern, allowing it to be rewritten as $y = \sqrt{x + \sqrt{y + y}}$.
This simplifies to $y = \sqrt{x + \sqrt{2y}}$.

Squaring both sides yields $y^2 = x + \sqrt{2y}$. Isolating the radical gives $y^2 - x = \sqrt{2y}$.
Squaring again produces $(y^2 - x)^2 = 2y$.

Differentiating implicitly with respect to $x$:

$2(y^2 - x)(2y\frac{dy}{dx} - 1) = 2\frac{dy}{dx}$

Dividing by 2 and expanding:

$2y(y^2 - x)\frac{dy}{dx} - (y^2 - x) = \frac{dy}{dx}$

Rearranging terms:

$\frac{dy}{dx}[2y(y^2 - x) - 1] = y^2 - x$

Solving for the derivative:

$\frac{dy}{dx} = \frac{y^2 - x}{2y^3 - 2xy - 1}$

This matches option C.