Question

Решите пожалуйста производную

Answer 1

$y=ln\frac{1+\sqrt{x^2+1}}{x}\\\\y'=\frac{x}{1+\sqrt{x^2+1}}\cdot \frac{\frac{2x}{2\sqrt{x^2+1}}\cdot x-(1+\sqrt{x^2+1})}{x^2}=\frac{1}{1+\sqrt{x^2+1}}\cdot \frac{x^2-\sqrt{x^2+1}-(x^2+1)}{x\cdot \sqrt{x^2+1}\cdot (1+\sqrt{x^2+1})}=\\\\=\frac{1}{1+\sqrt{x^2+1}}\cdot \frac{-(\sqrt{x^2+1}+1)}{x\cdot \sqrt{x^2+1}\cdot (1+\sqrt{x^2+1)}}=-\frac{1}{(1+\sqrt{x^2+1)\cdot x\cdot \sqrt{x^2+1}}}=-\frac{1}{x\cdot \sqrt{x^2+1}+x^3+x}$