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(D) Given the equation: $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}$We can rewrite the inner part as: $y=\sqrt{x+\sqrt{y+y}}$Squaring both si

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$\frac{y^2-x}{2 y^3-2 x y-1}$

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(D) Given the equation: $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}$We can rewrite the inner part as: $y=\sqrt{x+\sqrt{y+y}}$Squaring both sides: $y^2=x+\sqrt{2y}$Rearranging: $y^2-x=\sqrt{2y}$Squaring again: $(y^2-x)^2=2y$Expanding: $y^4-2xy^2+x^2=2y$Rearranging to implicit form: $y^4-2xy^2-2y+x^2=0$Differentiating with respect to $x$: $\frac{d}{dx}(y^4-2xy^2-2y+x^2) = 0$$4y^3 \frac{dy}{dx} - (2y^2 + 4xy \frac{dy}{dx}) - 2 \frac{dy}{dx} + 2x = 0$Grouping $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(4y^3 - 4xy - 2) = 2y^2 - 2x$Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2y^2-2x}{4y^3-4xy-2} = \frac{y^2-x}{2y^3-2xy-1}$

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

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