Differentiating from first principles

This question comes up frequently (not every year) in the Leaving Cert. When it comes up, you usually have to differentiate a quadratic, though you should practice differentiating linear functions (e.g. f(x) = 4x -1) also. It is usually worth 20 marks.

The way I approach it is to set up the derivative function as if you were finding the slope using the slope formula m = (y2-y1)/(x2-x1).
Today's example which we did in class was

3x²-4x+2 (I think)

x1 = x
y1 = 3x²-4x+2
x2 = x + h
y2 = 3(x+h)²-4(x+h)+2

If you plug these into the slope formula and multiply out the y2 and watch your signs with y1, you'll get

m =
3x²+3h² +6xh -4x-4h+2 - 3x²+4x-2
------------------------------------------------
x + h - x

= 3h² +6xh -4h
---------------
h

= 3h +6x -4

Now apply the Limit as h tends towards zero (you'll lose marks if you don't do this part properly)

Lim [f(x+h) - f(x)]/h = 6x -4
h->0

Also, write down somewhere either f(x) = 3x²-4x+2 and f'(x) = 6x -4

or
y = 3x²-4x+2

dy
-- = 6x -4
dx

It is worth spending time practicing and understanding this process. You only have to do it if you are specifically asked to differentiate "from first principles", but it is central to figuring out other topics in functions and differentiation, and it gives you an application of coordinate geometry and practice at some tricky algebra.

I have a worksheet that I didn't have time to distribute today that gives a worked example and practice of differentiating from first principles.

We'll look at the rules for differentiating (product, quotient and chain rules) tomorrow.
Test on Friday.