Найдите производную функцию

Решите задачу:

$\displaystyle y'= \frac{(-3x^4-4x^3)'*(-3x^2-3)-(-3x^4-4x^3)*(-3x^2-3)'}{(-3x^2-3)^2} =\\ \\ \\ = \frac{(12x^3+12x^2)(3x^2+3)+(3x^4+4x^3)*(-6x)}{9(x^2+1)^2} =\\ \\ \\ = \frac{36x^5+36x^3+36x^4+36x^2-18x^5-24x^4}{9(x^2+1)^2} =\\ \\ \\ = \frac{18x^5+12x^4+36x^3+36x^2}{9(x^2+1)^2} = \frac{9x^5+4x^4+12x^3+12x^2}{3(x^2+1)^2}$