maths derivatives

{Note: the ' symbol means d/dx}

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

Method 1: expand it and differentatiate them term by term

Method 2: Use the Chain Rule
y= (x-3)^2 * (x+2)^2 * (x+1) = [(x-3)^2] [(x+2)^2 * (x+1)]
y' = [(x-3)^2] ' [(x+2)^2 * (x+1)] + [(x-3)^2] [(x+2)^2 * (x+1)] '
=[(x-3)^2] ' [(x+2)^2 * (x+1)] + [(x-3)^2] [((x+2)^2)' * (x+1) + (x+2)^2 * (x+1)']
=2(x-3) [(x+2)^2 * (x+1)] + [(x-3)^2] [2(x+2) * (x+1) + (x+2)^2 * 1 ]
=(x-3) (x+2) (5 x^2+x-8) after factorization

Method 3:
y= (x-3)^2 * (x+2)^2 * (x+1)
ln y = 2ln (x-3) + 2ln (x+2) + ln(x+1)
Differentatiating with respect to x,
d(ln y)/dy * (dy/dx) = 2/(x-3) +2/(x+2) +1/(x+1)
(1/y)(dy/dx) = 2/(x-3) +2/(x+2) +1/(x+1)
dy/dx = y * [2/(x-3) +2/(x+2) +1/(x+1)]
= (x-3)^2 * (x+2)^2 * (x+1) * [2/(x-3) +2/(x+2) +1/(x+1)]
= (x-3) (x+2) (5 x^2+x-8) after simplifying

Using product rule : d(uvw) = uvdw + vwdu + uwdv
u = (x - 3)^2
v = (x + 2)^2
w = (x + 1)
so dy/dx = (x - 3)^2(x + 2)^2 d(x + 1)/dx + (x - 3)^2(x + 1) d(x + 2)^2/dx
+ (x + 2)^2(x + 1) d(x - 3)^2/dx
= (x - 3)^2(x + 2)^2 + 2(x - 3)^2(x + 1)(x + 2) + 2(x + 2)^2(x + 1) (x - 3)
= (x - 3)(x + 2)[(x - 3)(x + 2) + 2(x - 3)(x + 1) + 2(x + 2)(x + 1)]
= (x - 3)(x + 2)[x^2 - x - 6 + 2x^2 - 4x - 6 + 2x^2 + 6x + 4]
= (x - 3)(x + 2)(5x^2 + x - 8).