If y = sqrt {x + sqrt {x + sqrt {x + dots inf | vedclass

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(D) Given the infinite series $y = \sqrt {x + \sqrt {x + \sqrt {x + \dots \infty } } }$.Since the series is infinite,we can write it as $y = \sqrt {x

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

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(D) Given the infinite series $y = \sqrt {x + \sqrt {x + \sqrt {x + \dots \infty } } }$.Since the series is infinite,we can write it as $y = \sqrt {x + y}$.Squaring both sides,we get $y^2 = x + y$.Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(x + y)$.Using the chain rule,$2y \frac{dy}{dx} = 1 + \frac{dy}{dx}$.Rearranging the terms to isolate $\frac{dy}{dx}$,we get $2y \frac{dy}{dx} - \frac{dy}{dx} = 1$.$\frac{dy}{dx}(2y - 1) = 1$.Therefore,$\frac{dy}{dx} = \frac{1}{2y - 1}$.

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

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