We have to find the derivative of f(x) = (2x+1)^2+(2x-2)^2 + 2(4x^2-2x-2)

f'(x) = [(2x+1)^2+(2x-2)^2 + 2(4x^2-2x-2)]'

f'(x) = [(2x+1)^2]'+[(2x-2)^2]' + [2(4x^2-2x-2)]'

Use the chain rule.

f'(x) = 2(2x + 1)*2 + 2*(2x - 2)*2 + 2*(8x - 2)

=> 8x + 4 + 8x - 8 + 16x - 4

=> 32x - 8

**The derivative of the function is 32x - 8**

We notice that the 3rd term is the result of the product (2x+1)(2x-2) = (4x^2 - 2x - 2).

If we'll note (2x+1) by a and (2x-2) by b, and we'll re-write the equation, we'll get a perfect square:

y=a^2 + 2ab + b^2

y = (a+b)^2

f(x) = y = (2x+1+2x-2)^2

We'll combine like terms:

y = (4x-1)^2

Now, we'll differentiate both sides, with respect to x:

dy/dx = 2*(4x-1)*(4x-1)'

dy/dx = 2*(4x-1)*4

dy/dx = 8*(4x-1)

We'll remove the brackets:

dy/dx = 32x - 8

**The first derivative of the given function is f'(x) = 32x - 8.**