If y=sqrt{x+sqrt{x+sqrt{x+ldots infty}}} ,th | vedclass

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(D) Given that,$ y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} $Since the expression under the square root repeats infinitely,we can write:$ y=\sqrt{x+y}

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$ \frac{1}{2 y-1} $

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(D) Given that,$ y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} $Since the expression under the square root repeats infinitely,we can write:$ y=\sqrt{x+y} $Squaring both sides,we get:$ y^{2}=x+y $Differentiating both sides with respect to $ x $:$ \frac{d}{dx}(y^{2}) = \frac{d}{dx}(x+y) $$ 2y \frac{dy}{dx} = 1 + \frac{dy}{dx} $Rearranging the terms to solve for $ \frac{dy}{dx} $:$ 2y \frac{dy}{dx} - \frac{dy}{dx} = 1 $$ \frac{dy}{dx}(2y - 1) = 1 $$ \frac{dy}{dx} = \frac{1}{2y - 1} $

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

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