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Let $x = \tan\theta$. Since $x \in (0,1)$, we have $\theta \in (0, \frac{\pi}{4})$.
Substitute $x = \tan\theta$ into the expression for $y$:\r\n$$y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$$\r\nUsing the trigonometric identity $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$, we get:\r\n$$y = \cos^{-1}(\cos 2\theta)$$\r\nSince $\theta \in (0, \frac{\pi}{4})$, we have $2\theta \in (0, \frac{\pi}{2})$. Therefore, $\cos^{-1}(\cos 2\theta) = 2\theta$. So,\r\n$$y = 2\theta$$
Since $x = \tan\theta$, we have $\theta = \tan^{-1}x$. Therefore,\r\n$$y = 2\tan^{-1}x$$
Differentiate $y$ with respect to $x$:\r\n$$\frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}x) = 2\frac{d}{dx}(\tan^{-1}x)$$\r\nUsing the derivative of $\tan^{-1}x$, which is $\frac{1}{1+x^2}$, we get:\r\n$$\frac{dy}{dx} = 2\left(\frac{1}{1+x^2}\right) = \frac{2}{1+x^2}$$
Final Answer: $\frac{2}{1+x^2}$
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