Question

СРОЧНО СРОЧНО СРОЧНО ...

Answer 1

Решение представлено во вложении

Answer 2

$f(x)=\frac{2x}{3-\sqrt{x^2+1}}\\\\ f'(x)=\frac{2*(3-\sqrt{x^2+1})-2x*[0-\frac{2x}{2\sqrt{x^2+1}}]}{(3-\sqrt{x^2+1})^2}=\\\\ =\frac{2*(3-\sqrt{x^2+1})+\frac{2x^2}{\sqrt{x^2+1}}}{(3-\sqrt{x^2+1})^2}\\\\ f'(1)=\frac{2*(3-\sqrt{2})+\frac{2}{\sqrt{2}}}{(3-\sqrt{2})^2}$

$f(x)=ln(\frac{2x-1}{2x+1})\\\\ f'(x)=\frac{2x+1}{2x-1}*\frac{2*(2x+1)-(2x-1)*2}{(2x+1)^2}=\\\\ =\frac{4}{4x^2-1}\\\\ f'(0)=\frac{4}{-1}=-4$