I have always done formal derivatives using a delta x which represents the change in the x values of two points along the curve. This definition literally defines rise over run (slope) as that distance becomes nonexistant. lim as DelX -->0 of [f(x+Delx)-f(x)]/[Delx]. This works on a function like 2x^2 as follows. [2(x+Delx)^2-2x^2]/Delx ... [2(x^2 +2xDelx +Delx^2)-2x^2]/Delx .. [4xDelx + 2Delx^2]/Delx ... 4x + 2 Delx {Delx-->0} = 4x. If we were to plug in at say an x value of 3 (y value 18) we find that the slope is 12.

But I was tutoring some precalculus the other day and they were doing slopes. (essentially derivatives) The only difference was their method. There formula was like this lim as x-->c of [f(x)-f(c)]/[x-c] In this form, the 2x^2 would behave like this. We would assume a point such as the previous (3,18) and continue as such: [2x^2-18]/[x-3] ... 2(x+3)(x-3)/(x-3) ... 2(x+3) lim x-->c = 12 .

We found that in both cases the slope was 12 though the first one came out to a formula 4x = m and the second as 2(x+3) = m at x=3. And this brings me to the main flaw in the precalc method. The point must be defined beforehand. 2(x+3) only works at x=3..... anywhere else such as at 4 and it gives us a slope of 14 (should be 16) so that for a slope at two different points would require two different limits of essentially the same problem.The formal definition that I did first can be done once and ANY point can be plugged into the resulting formula to give the slope. So you pre-calc students can keep that x approaches c thing to yourselves! :(