Question

Продифференцировать, , ! y=(1-x^3)/(1+x^3) y'=?

Answer 1

$y'_x(x)=( \frac{1-x^3}{1+x^3} )'=((1-x^3)(1+x^3)^{-1})'=$
$=(1-x^3)'(1+x^3)^{-1}+(1-x^3)((1+x^3)^{-1})'=$
$=(-3x^2)(1+x^3)^{-1}+(1-x^3)(-1)(1+x^3)^{-2}(1+x^3)'=$
$= -\frac{3x^2}{1+x^3} -(1-x^3)(1+x^3)^{-2}(3x^2)=$
$= -\frac{3x^2}{1+x^3}- \frac{3x^2(1-x^3)}{(1+x^3)^2} =- \frac{3x^2(1+x^3+1-x^3)}{(1+x^3)^2} =- \frac{6x^2}{(1+x^3)^2}$