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

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(D) Given the equation: $y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}$Squaring both sides,we get: $y^2 = x + \sqrt{x + \sqrt{x + \ldots}}$Since the exp

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

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

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

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