Gotta use the multiplication rule for that problem. I was taught it as U`V + UV`, aka, the derivative of the first term, times the second, plus the first term time the derivative of the second. However, since the second term is in brackets, we need to use the chain rule to determine it's derivative. That gives us 2x(x-3) + X^2(1)(x-3)^0(1), which can be simplified down to 2x^2 - 3x + x^2, then to 3x^2 - 3x, then further down to x^2 - x. We didn't have to use the chain rule in this case, as anything to the power of 0 is 1, but its good practice to do so incase you come across something in brackets to the power of some other number.

my roomate and i may be wrong but shouldent it be That gives us 2x(x-3) + X^2(1)(x-3)^0(1), which can be simplified down to 2x^2 - 6x + x^2, then to 3x^2 - 6x, then further down to x^2 - 2x. because 2x(x-3) simplifyes to 2x^2-6x