1. **Problem statement:** Differentiate $$\arctan \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$$ with respect to $$\arccos x^{2}$$. 2. **Rewrite the function:** Let $$f(x) = \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}} + \sqrt{1-x^{2}}}$$ and define $$u = \arctan f(x)$$ and $$v = \arccos x^{2}$$. 3. **Goal:** Find $$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}$$. 4. **Differentiate $$u = \arctan f(x)$$:** $$\frac{du}{dx} = \frac{1}{1 + f(x)^2} \cdot f'(x)$$. 5. **Simplify $$f(x)$$:** Multiply numerator and denominator by $$\sqrt{1+x^{2}} - \sqrt{1-x^{2}}$$: $$f(x) = \frac{(\sqrt{1+x^{2}} - \sqrt{1-x^{2}})^2}{(\sqrt{1+x^{2}})^2 - (\sqrt{1-x^{2}})^2} = \frac{(\sqrt{1+x^{2}} - \sqrt{1-x^{2}})^2}{(1+x^{2}) - (1 - x^{2})} = \frac{(\sqrt{1+x^{2}} - \sqrt{1-x^{2}})^2}{2x^{2}}$$. 6. **Expand numerator:** $$(\sqrt{1+x^{2}})^2 - 2\sqrt{1+x^{2}}\sqrt{1-x^{2}} + (\sqrt{1-x^{2}})^2 = (1+x^{2}) - 2\sqrt{(1+x^{2})(1-x^{2})} + (1 - x^{2}) = 2 - 2\sqrt{1 - x^{4}}$$. 7. **So,** $$f(x) = \frac{2 - 2\sqrt{1 - x^{4}}}{2x^{2}} = \frac{1 - \sqrt{1 - x^{4}}}{x^{2}}$$. 8. **Differentiate $$f(x)$$:** $$f'(x) = \frac{d}{dx} \left( \frac{1 - \sqrt{1 - x^{4}}}{x^{2}} \right) = \frac{(0 - \frac{-4x^{3}}{2\sqrt{1 - x^{4}}}) x^{2} - (1 - \sqrt{1 - x^{4}}) 2x}{x^{4}}$$ $$= \frac{\frac{4x^{3}}{2\sqrt{1 - x^{4}}} x^{2} - 2x + 2x \sqrt{1 - x^{4}}}{x^{4}} = \frac{2x^{5}/\sqrt{1 - x^{4}} - 2x + 2x \sqrt{1 - x^{4}}}{x^{4}}$$. 9. **Simplify numerator:** $$2x^{5}/\sqrt{1 - x^{4}} - 2x + 2x \sqrt{1 - x^{4}} = 2x \left( \frac{x^{4}}{\sqrt{1 - x^{4}}} - 1 + \sqrt{1 - x^{4}} \right)$$. 10. **Therefore,** $$f'(x) = \frac{2x \left( \frac{x^{4}}{\sqrt{1 - x^{4}}} - 1 + \sqrt{1 - x^{4}} \right)}{x^{4}} = \frac{2 \left( \frac{x^{4}}{\sqrt{1 - x^{4}}} - 1 + \sqrt{1 - x^{4}} \right)}{x^{3}}$$. 11. **Calculate $$\frac{du}{dx}$$:** $$\frac{du}{dx} = \frac{1}{1 + f(x)^2} f'(x)$$. 12. **Differentiate $$v = \arccos x^{2}$$:** $$\frac{dv}{dx} = -\frac{1}{\sqrt{1 - (x^{2})^{2}}} \cdot 2x = -\frac{2x}{\sqrt{1 - x^{4}}}$$. 13. **Final derivative:** $$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{1}{1 + f(x)^2} f'(x)}{-\frac{2x}{\sqrt{1 - x^{4}}}} = - \frac{\sqrt{1 - x^{4}}}{2x (1 + f(x)^2)} f'(x)$$. 14. **Summary:** The derivative is $$\boxed{\frac{d}{d(\arccos x^{2})} \arctan \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}} + \sqrt{1-x^{2}}} = - \frac{\sqrt{1 - x^{4}}}{2x (1 + f(x)^2)} f'(x)}$$ where $$f(x) = \frac{1 - \sqrt{1 - x^{4}}}{x^{2}}$$ and $$f'(x)$$ as above. --- **Note:** The second problem is not solved as per instructions to solve only the first problem.