Question

НАЙДИТЕ ПОЖАЛУЙСТА ПРОИЗВОДНУЮ​

Answer 1

Ответ:

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