The point of a derivative, in elementary
calculus at least, is to find the
gradient of a function at any particular point P. Before we learn about calculus, the problem is usually solved by drawing an approximation to the
tangent at the point P, forming a right-angled
triangle and dividing the length of the
vertical side by the length of the
horizontal side. This gives the gradient of the
hypotenuse of the triangle and hence an approximation to the gradient of the function at the point P.
If we think about the triangle drawing idea above and bring it to a limit, so that we're drawing an arbitrarily small triangle, then we approach the true gradient at the point. This lets us derive the
formula for a derivative from
first principles.
Say we have the function f and we want to find the gradient at the point x
1. Then we 'draw' a horizontal line from the point (x
1, f(x
1)) to the point (x
1 + δx, f(x
1)) and a vertical line from the point (x
1 + δx, f(x
1)) to the point (x
1 + δx, f(x
1 + δx)), where δx is a small change in the x-
coordinate.
Now, dividing the length of the vertical line by the length of the horizontal line, and cancelling, gives us the approximation to the gradient at the point x
1:
f(x1 + δx) - f(x1)
---------------------
δx
Now, this should give us the exact gradient if we let δx go to zero, so the
formula we eventually obtain for the gradient of a function f at a point x
1 is:
f(x1 + δx) - f(x1)
f|(x1) = limδx->0 ---------------------
δx
Now for a quick
example. Anyone who's taken even the most elementary of elementary calculus courses knows that the derivative of x
n is nx
n-1. Pumping f(x) = x
n into the formula
derived above gives:
(x1 + δx)n - x1n
f|(x1) = limδx->0 ---------------------
δx
Now, by the
binomial theorem:
(x
1 + δx)
n = x
1n + nx
1n-1δx + (n(n-1)/2)x
1n-2δx
2 + o(δx
3)
where o(δx3) means terms where δx is present in at least the third power.
So:
x1n + nx1n-1δx + (n(n-1)/2)x1n-2δx2 + o(δx3) - x1n
f|(x1) = limδx->0 ----------------------------------------------------
δx
nx1n-1δx + (n(n-1)/2)x1n-2δx2 + o(δx3)
= limδx->0 --------------------------------------
δx
nx1n-1δx (n(n-1)/2)x1n-2δx2 o(δx3)
= limδx->0 ------- + ----------------- + -------
δx δx δx
lim
δx->0 (δx
2/δx) and lim
δx->0 (o(δx
3)/δx) are both zero, so, after cancelling δx's, we are left with:
f|(x1) = nx1n-1
which is exactly the result we were looking for. Similar arguments (using expansions and cancelling) can be used to derive the derivatives of lots of functions including the
trigonometric functions.
I've used the <pre> tag lots and lots and it looks fine on my system. If it looks all funny, please /msg me.