Question

СРОЧНО СРОЧНО СРОЧНО ...

Answer 1

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

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