Áp dụng công thức tính đạo hàm ` (u/v)' = (u'v - uv')/(v^2)`
Đặt ` u = (x+1)^3 ; v = (x-1)^2`
` => u' = (x+1)^3' = 3(x+1)^2`
` v' = (x-1)^2' = 2(x-1)`
` => y' = (((x+1)^3)/((x-1)^2))' = (3(x+1)^2*(x-1)^2 - (x+1)^3*2*(x-1))/((x-1)^4)`
` = (3(x-1)^2(x+1)^2 - (x^3 +3x^2 +3x +1)*2*(x-1))/((x-1)^4)`
` = (3(x-1)^2(x+1)^2 - (2x^3 + 6x^2 +6x +2)(x-1))/(x-1)^4`
` = (3(x-1)(x+1)^2 - 2x^3 -6x^2 -6x -2)/((x-1)^3)`
` = (3*(x^2-1)(x+1) - 2x^3 -6x^2 -6x -2)/((x-1)^3)`
` = (3x^3 + 3x^2 - 3x-3 -2x^3 -6x^2 -6x -2)/((x-1)^3)`
` = (x^3 - 3x^2 - 9x - 5)/((x-1)^3)`
