How do you differentiate this problem?!

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

I did chain rule for first quantity, multiplied the (3x-2) and (x-1) together, and then used the product rule, but I keep getting crazy answers

Sometimes it's less error-prone to just expand the terms:

y = (2x + 3)²(3x - 2)(x - 1)
y = (4x² + 12x + 9)(3x² - 5x + 2)
y = 12x^4 - 20x³ + 8x² + 36x³ - 60x² + 24x + 27x² - 45x + 18
y = 12x^4 - 20x³ + 36x³ + 8x² - 60x² + 27x² + 24x - 45x +18
y = 12x^4 + 16x³ - 25x² - 21x +18

y' = 48x³ + 48x² - 50x - 21
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I would multiply everything out first and then differentiate els its a mess trying to do the product rule that many times. If you dont want to multiply EVERYTHING then just expand the squared term and FOIL the 3x-2 and x-1 terms out; then do the product rule.

i got: product rule on (4x^2+12x+9)(3x^2-5x+2) which gives (8x+12)(3x^2-5x+2)+(4x^2+12x+9)(6x-5).

Given: f(x) = (2x + 3)²(3x - 2)(x - 1)

let g(x) = (2x + 3)²
let h(x) = (3x - 2)(x - 1)

This makes:

f(x) = g(x)h(x)

so that we may use the product rule to compute f'(x):

f'(x) = g'(x)h(x) + g(x)h'(x)

To compute g'(x) we must use the chain rule:

g'(u(x)) = {g'(u)}{u'(x)}

let u(x) = 2x + 3, then g(u) = u², g'(u) = 2u, and u'(x) = 2

g'(u(x)) = {2u}{2} = 4(2x + 3) = 8x + 12

To compute h'(x) we must, again, use the product rule:

let p(x) = 3x - 2
let q(x) = x - 1

h(x) = p(x)q(x)

h'(x) = p'(x)q(x) + p(x)q'(x)

p'(x) = 3
q'(x) = 1

h'(x) = 3(x - 1) + 1(3x - 2) = 3x - 3 + 3x - 2 = 6x - 5

Now that we have computed g'(x) and h'(x), we may to the product rule for f'(x):

f'(x) = g'(x)h(x) + g(x)h'(x)

f'(x) = {8x + 12}{(3x - 2)(x - 1)} + {(2x + 2)²}{6x - 5}