If $ y = \tan^{-1}(\sqrt{x^2 - 1}) $, then the ratio $ \frac{d^2y}{dx^2} : \frac{dy}{dx} = \ldots $.

Question

$ \frac{1 + 2x^2}{x(x^2 + 1)} $

$ \frac{1 - 2x^2}{x(x^2 + 1)} $

$ \frac{1 + 2x^2}{x(x^2 - 1)} $

$ \frac{1 - 2x^2}{x(x^2 - 1)} $

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