Question

Найти вторую производную

Answer 1

y' = (2x(x+2) - x^2)/(x+2)^2 = (2x^2 + 4x - x^2)/(x+2)^2 = (x^2 +4x)/(x+2)^2;

y'' = ((2x + 4)(x^2 + 4x + 4) - (x^2 + 4x)(2x + 4))/(x+2)^4 = (2x + 4)(x^2 + 4x + 4 - x^2 - 4x)/(x+2)^4 = 8(x+2)/(x+2)^4 = 8/(x+2)^3.

Answer 2

$y=frac{x^2}{x+2}\\y'=frac{2x(x+2)-x^2cdot 1}{(x+2)^2}=frac{x^2+4x}{(x+2)^2}=frac{x(x+4)}{(x+2)^2}\\y''=frac{(2x+4)(x+2)^2-(x^2+4x)cdot 2(x+2)}{(x+2)^4}=frac{(x+2)cdot (, (2x+4)(x+2)-2cdot (x^2+4x), )}{(x+2)^4}=\\=frac{2x^2+4x+4x+8-2x^2-8x}{(x+2)^3}=frac{8}{(x+2)^3}$