Find \( \frac{d y}{d x} \) for \( y=\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right\} \)

To find \( \frac{dy}{dx} \) for the function \[ y = \tan^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right), \] we use the chain rule and the quotient rule. After differentiating, the derivative is: \[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right)^2} \cdot \frac{\frac{x}{\sqrt{1+x^{2}}} - \frac{x}{\sqrt{1-x^{2}}}}{\left(\sqrt{1+x^{2}} - \sqrt{1-x^{2}}\right)^2}. \] Simplifying further, we get: \[ \frac{dy}{dx} = \frac{x}{1 - x^{2}}. \]