Re: what "REALLY" is derivative?
- From: "Randy Poe" <poespam-trap@xxxxxxxxx>
- Date: 13 Dec 2005 13:27:59 -0800
Luke Wu wrote:
> Hi,
>
> Back in grade 11 when learning about parabolas, a teacher told me that
> there are lines that can touch the parabola only at one point, for
> every point on the curve (tangents), and that we needed to wait for
> more advanced courses to find the equation of these lines.
>
> I thought this was nonsense, because if two curves (parabola and line)
> intersect, then it's easy to solve for the slope of the line (because
> we knew how to solve the intersection of two lines- "this can't be much
> different" i thougth).. this is what I did:
>
> equation of parabola: y = ax^2 + bx + c
>
> point slope equation of line: y - y1 = m(x - x1)
>
> what I did was substitute the point (q , aq^2 + bq + c) for (x1,y1) in
> the slope equation
>
>
> y - aq^2 - bq - c = m(x - q)
>
> then substitute the y = aq^2 + bq + c
>
> ax^2 + bx + c - aq^2 - bq - c = m(x - q)
>
> a(x^2 - q^2) + b(x - q)= m(x - q)
>
> (x-q) [ a(x+q) + b ] = m(x - q)
>
> THEN I DID THE UNTHINKABLE and divided both sides by (x-q)
>
> m = a(x+q) + b and that is the slope of a line touching parabola at x
> = q
There's a lot wrong with that derivation, but in the end what
you did was divide delta-y by delta-x. Since a term (x-q) appears
on both sides, you get m = a(x+q) + b for all x NOT EQUAL TO q,
and therefore in the limit as x->q, m = 2aq + b.
> I had no clue what a limit was, or a derivative, but I used the ideas
> learned about the intersection of two lines, and used it for the
> intersection of a parbola and a line (and just ASSUMED that x-q != 0)
>
> I then tried this for other curves successfully, and when I started
> learning calculus (2 years later), I couldn't get past the fact that
> the limit seemed to me (for reasons shown above) as an excuse to allow
> us to divide f(x) - f(a) by x - a.
Evaluating a limit as x->a is not the same as evaluating at
x=a, and indeed the two are only the same when a function is
continuous (that's a definition of continuity).
Since it's perfectly valid to divide f(x)-f(a) by (x-a) when x
is not equal to a, this gives you a perfectly good sequence
of values as x approaches a, and if the sequence has a limit,
then that limit is the derivative at a.
> How is there always a x-a (for us to cancel out) within f(x) - f(a) if
> f is differentiable at 'a'.
There isn't, in general. What about the derivative of sin(x)?
However, Taylor's Theorem always applies. Have you seen
Taylor series?
- Randy
.
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