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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.
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