Question

Помогите найти производную.

Answer 1

$y' = (\ln(\frac{\sqrt{x^2 + 1} + x\sqrt{2}}{\sqrt{x^2 + 1} - x\sqrt{2}}))' = \frac{\sqrt{x^2 + 1} - x\sqrt{2}}{\sqrt{x^2 + 1} + x\sqrt{2}}\cdot \frac{(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{2})(\sqrt{x^2 + 1} - x\sqrt{2}) - (\sqrt{x^2 + 1} + x\sqrt{2})(\frac{x}{\sqrt{x^2 + 1}} - \sqrt{2})}{(\sqrt{x^2 + 1} - x\sqrt{2})^2}$

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