If y = sqrt{x + sqrt{y + sqrt{x + sqrt{y + do | vedclass

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(A) Given the equation $y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}}$.Squaring both sides,we get $y^2 = x + \sqrt{y + \sqrt{x + \sqrt{

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

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(A) Given the equation $y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}}$.Squaring both sides,we get $y^2 = x + \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}$.Let $u = \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}$. Then $y^2 = x + u$.Squaring $u$,we get $u^2 = y + \sqrt{x + \sqrt{y + \dots \infty}} = y + y = 2y$.Thus,$u = \sqrt{2y}$.Substituting $u$ back into the equation for $y^2$,we have $y^2 = x + \sqrt{2y}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = 1 + \frac{1}{2\sqrt{2y}} \cdot 2 \frac{dy}{dx}$.$2y \frac{dy}{dx} = 1 + \frac{1}{\sqrt{2y}} \frac{dy}{dx}$.$\frac{dy}{dx} (2y - \frac{1}{\sqrt{2y}}) = 1$.$\frac{dy}{dx} (\frac{2y\sqrt{2y} - 1}{\sqrt{2y}}) = 1$.$\frac{dy}{dx} = \frac{\sqrt{2y}}{2y\sqrt{2y} - 1}$.Alternatively,from $y^2 - x = \sqrt{2y}$,squaring gives $(y^2 - x)^2 = 2y$.Differentiating: $2(y^2 - x)(2y \frac{dy}{dx} - 1) = 2 \frac{dy}{dx}$.$(y^2 - x)(2y \frac{dy}{dx} - 1) = \frac{dy}{dx}$.$2y(y^2 - x) \frac{dy}{dx} - (y^2 - x) = \frac{dy}{dx}$.$\frac{dy}{dx} (2y^3 - 2xy - 1) = y^2 - x$.$\frac{dy}{dx} = \frac{y^2 - x}{2y^3 - 2xy - 1}$.

If $y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \dots \infty}}}}$,then $\frac{dy}{dx} = $

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