Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3

A=delta (dont know how to do delta)

im doing it using this method
( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x )

i simplify and get Ax^2 + 3xAx + 3x^2 - 6x