[QUOTE=Jinx]
Seems trivial, perhaps, but when (presumably) Newton invented Calculus, for example…how did he prove that every derivative is the instanteous slope of a curve …
[/QUOTE]
The derivative is pretty much the instantaneous slope of a curve by definition.
Take any two points on a set of cartesian coordinates, defined by the ordered pairs (x1, y1) and (x2, y2).
The slope of the straight line that connects these two points is given by
m = deltaY/deltaX = (y2-y1)/(x2-x1)
Now, assume that the two points are not just randomly selected points, but in fact lie on the graph of some function. We can see a number of things:
[ul]
[li]y1 is f(x1)[/li][li]x2 is x1 plus some interval h, so x2 is x1+h[/li][li]Therefore, y2 = f(x2) = f(x1+h)[/li][/ul]
So if, to drop our “subscripts” we let
x1 = x,
y1 = f(x)
x2 = x+h
and
y2 = f(x+h)
our points are now (x, x+h) and (f(x), f(x+h)).
Plugging these into the slope formula, we get
m = (f(x+h) - f(x)) / ((x+h) - x)
or (f(x+h) - f(x)) / h
So this slope represents the slope of any straight line connecting any two points that lie on the graph of the function.
If we want to figure out which way the graph is “going” at any particular point, we want the straight-line slope for two points on the graph that lie so close together that the distance is negligible. In other words, we want h, the interval that separates the x coordinates of the points, to approach zero, on the realization that, being a function, the y values would similarly converge in most cases, bringing the points on the graph vanishingly close together.
So we define this instantaneous slope as
d f(x)/dx = lim(h–>0) of (f(x+h) - f(x)) / h
We call that the derivative, and voila!