Y=3x^2+6x-7/(x+1)^2 show that dy/dx = 20/(x+1)^3

Apply the quotient rule  
d(uv)  =    v du/dx - u dv/dx
  dx          
                         v^2

 du/dx  =  6x + 6
 dv/dx   =  2(x + 1)
 v^2 = (x + 1)^4   so we have

dy dx =  (x + 1)^2 .* (6x + 6) - (3x^2 + 6x - 7) * 2(x + 1)   /  (x + 1)^4

         =   (x + 1) [(6(x + 1)^2 - 2(3x^2 + 6x - 7) ]  / (x + 1)^4

         =  (x^2 + 12x +6 - 6x^2  - 12x + 14 ) / (x + 1)^3  

        =  20 / (x + 1)^3   Answer

Where are the