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If $$y = \tan^{-1}\frac{2x}{1 - x^2}$$, then $$\frac{dy}{dx}$$ is
$$\frac{2}{1 + x^2}$$
$$\frac{2}{1 - x^2}$$
$$\frac{1}{1 + x^2}$$
$$\frac{2x}{\sqrt{1 + x^2}}$$
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