Re: what "REALLY" is derivative?
- From: "Narasimham" <mathma18@xxxxxxxxxxx>
- Date: 13 Dec 2005 14:06:03 -0800
> How is there always a x-a (for us to cancel out) within f(x) - f(a) if
> f is differentiable at 'a'. Does anyone know how this really works?
By remainder theorem if F(x,a) vanishes at x = a, then (x-a) is a
factor for F(x,a). Or if you divide F(x,a) by (x-a) remainder is zero.
When [f(x) - f(a) ] /(x-a) assumes the form 0/0 it does not go up in
smoke but tends to the limit of derivative evaluated at that point.
This has been called the fundamental theorem of calculus. E.g., slope
of ax^2 + bx + c is 2*a*x + b at any x. At x =q this is 2 a q +b
which is your m = a (x+q) + b at x =q .
> and that is the slope of a line touching parabola at x = q.
NO. This is the slope of the secant cutting the parabola at two
distinctly separate points. Only when the q value comes near to x, the
secant becomes a tangent which you can now , if you will, still see it
as a secant at two coincident points.
To convince yourself draw a tangent to y = x^2 at x= 1. Then join two
points on the parabola whose x- coordinates are (1and 2), (1and 1.8),
(1 and 1.4),(1 and 1.1) etc., confirming the calculated slope of secant
for each pair. HTH
.
- References:
- what "REALLY" is derivative?
- From: Luke Wu
- what "REALLY" is derivative?
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