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(x_{1} + δx) - f(x_{1})
---------------------
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(x_{1} + δx) - f(x_{1})
f^{|}(x_{1}) = lim_{δx->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:

(x_{1} + δx)^{n} - x_{1}^{n}
f^{|}(x_{1}) = lim_{δx->0} ---------------------
δx

Now, by the

binomial theorem:
(x

_{1} + δx)

^{n} = x

_{1}^{n} + nx

_{1}^{n-1}δx + (n(n-1)/2)x

_{1}^{n-2}δx

^{2} + o(δx

^{3})

where o(δx^{3}) means terms where δx is present in at least the third power.

So:

x_{1}^{n} + nx_{1}^{n-1}δx + (n(n-1)/2)x_{1}^{n-2}δx^{2} + o(δx^{3}) - x_{1}^{n}
f^{|}(x_{1}) = lim_{δx->0} ----------------------------------------------------
δx
nx_{1}^{n-1}δx + (n(n-1)/2)x_{1}^{n-2}δx^{2} + o(δx^{3})
= lim_{δx->0} --------------------------------------
δx
nx_{1}^{n-1}δx (n(n-1)/2)x_{1}^{n-2}δx^{2} o(δx^{3})
= lim_{δx->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^{|}(x_{1}) = nx_{1}^{n-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.