Differentiation help

what is the derivative of
(x^3+3x^2)/(x^2+1)^2

I've tried using the quotient rule but that didnt work out.

(x^3+3x^2)/(x^2+1)^2

[ (x^2 +1)^2 (3x^2 + 6x) - (x^3 + 3x^2) (2) (x^2 +1) (2x) ] / (x^2 + 1)^4

=> [ (x^2 +1)^2 (3x^2 + 6x) - 4x (x^3 + 3x^2) (x^2 +1) ] / (x^2 + 1)^4

=> (x^2 + 1) [ (3x^2 + 6x)(x^2 +1)- 4x (x^3 + 3x^2 ) ] / (x^2 + 1)^4

=> [ (3x^2 + 6x)(x^2 +1) - 4x (x^3 + 3x^2) ] / (x^2 + 1)^3

=> (3x^4 + 6x^3 + 3x^2 + 6x - 4x^4 - 12x^3) / (x^2 + 1)^3

=> (-x^4 - 6x^3 + 3x^2 + 6x) / (x^2 +1)^3 . . .answer

edit: i want to make this clear. . .

(x^3+3x^2)(2)(x^2+1)(2x)

from: [ (u)(Dv) - (v) (Du) ] / u^2

where: u = (x^2+1)^2, . . .and v = (x^3+3x^2)

so, (v)(Du) ---> (x^3 + 3x^2) (2) (x^2 +1) (2x)

Quotient Rule: d(u/v) / dx = (v*du/dx - u*dv /dx)/ v^2

here, u = x^3 + 3x^2 =>du/dx = (3x^2 + 6x)
v = (x^2 + 1)^2 => dv/dx = 2*(x^2 + 1)*2x = 4x(x^2 + 1)

So d(u/v) / dx = [(x^2 + 1)^2 * (3x^2 + 6x) - (x^3 + 3x^2)*4x(x^2 + 1)] / (x^2 + 1)^4
You can probably simplify it further by expanding the numerator

y = x³ +3x² / (x²+1)²

y ' = (x²+1)² *(3x² +6x) -(x³ +3x²)(2)(x²+1)*2x / (x²+1)⁴

y ' = (x² +1) [ (x²+1)(3x²+6x) -4x (x³+3x²)] / (x²+1)⁴

y ' = [ (x²+1)(3x²+6x) -4x (x³+3x²)] / (x²+1)³

y = (x^3+3x^2)/(x^2+1)^2
= (x^3+3x^2) * (x^2+1)^(-2)

dy/dx = (3x^2 + 6x) * (x^2+1)^(-2) + (x^3+3x^2) * (-2) * (x^2+1)^(-3) * (2x)

= [3(x^2 + 2x) / (x^2 + 1)^2 ] - [4x(x^3 + 3x^2) / (x^2 + 1)^3]

yeah you need the quotient rule *sigh*

((x^2+1)^2(3x^2+6x) - (x^3+3x^2) 4x(x^2+1)) / (x^2+1)^4


((x^2+1)(3x^2+6x) - 4x(x^3+3x^2)(x^2+1)) / (x^2+1)^3