If sqrt{y-sqrt{y-sqrt{y-ldots infty}}} = sqrt | vedclass

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(C) Let $u = \sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}}$ and $v = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$.Given $u = v$.For $u$,we have $u = \sqrt{y-u

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$\frac{y-x+1}{y-x-1}$

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(C) Let $u = \sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}}$ and $v = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$.Given $u = v$.For $u$,we have $u = \sqrt{y-u} \implies u^2 = y-u \implies y = u^2+u$.For $v$,we have $v = \sqrt{x+v} \implies v^2 = x+v \implies x = v^2-v$.Since $u = v$,we can write $y = u^2+u$ and $x = u^2-u$.Now,differentiate $y$ and $x$ with respect to $u$:$\frac{dy}{du} = 2u+1$ and $\frac{dx}{du} = 2u-1$.Using the chain rule,$\frac{dy}{dx} = \frac{dy/du}{dx/du} = \frac{2u+1}{2u-1}$.From $y = u^2+u$,we have $u^2+u-y = 0$. Solving for $u$ using the quadratic formula: $u = \frac{-1 \pm \sqrt{1+4y}}{2}$.Alternatively,express the result in terms of $x$ and $y$:Since $y-x = (u^2+u) - (u^2-u) = 2u$,we have $u = \frac{y-x}{2}$.Substituting $u$ into $\frac{dy}{dx} = \frac{2u+1}{2u-1}$:$\frac{dy}{dx} = \frac{2(\frac{y-x}{2})+1}{2(\frac{y-x}{2})-1} = \frac{y-x+1}{y-x-1}$.

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

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