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

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(D) Let $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} = z$. Since the expressions are infinite,we can write: $y

Vedclass · Accepted Answer

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

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(D) Let $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} = z$. Since the expressions are infinite,we can write: $y - z = z^2 \Rightarrow y = z^2 + z$ $x + z = z^2 \Rightarrow x = z^2 - z$ Now,differentiate both with respect to $z$: $\frac{dy}{dz} = 2z + 1$ $\frac{dx}{dz} = 2z - 1$ Using the chain rule,$\frac{dy}{dx} = \frac{dy/dz}{dx/dz} = \frac{2z+1}{2z-1}$. From the equations $y = z^2 + z$ and $x = z^2 - z$,we subtract them: $y - x = (z^2 + z) - (z^2 - z) = 2z$. Substituting $2z = y - x$ into the derivative expression: $\frac{dy}{dx} = \frac{(y-x) + 1}{(y-x) - 1} = \frac{y-x+1}{y-x-1}$.

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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