If $ y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} $, then find $ \frac{dy}{dx} $.

Question

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

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

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

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

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