TITLE: Defining the derivative without limits
QUESTION [25 upvotes]: These days, the standard way to present differential calculus is by introducing the Cauchy-Weierstrass definition of the limit. One then defines the derivative as a limit, proves results like the Leibniz and chain rules, and uses this machinery to differentiate some simple functions such as polynomials. The purpose of my question is to see what creative alternatives people can describe to this approach. The nature of the question is that there is not going to be a single best answer. I have several methods that I've collected which I'll put in as answers to my own question.
It's not reasonable to expect answers to include an entire introductory textbook treatment of differentiation, nor would anyone want to read answers that were that lengthy. A sketch is fine. Lack of rigor is fine. Well known notation and terminology can be assumed. It would be nice to develop things to the point where one can differentiate a polynomial, since that would help to illustrate how your method works and demonstrate that it's usable. For this purpose, it suffices to prove that if $n>0$ is an integer, the derivative of $x^n$ equals $0$ at $0$ and equals $n$ at $1$; the result at other nonzero values of $x$ follows by scaling. Doing this for $n=2$ is fine if the generalization to $n>2$ is obvious.

REPLY [2 votes]: The question is, given a graph $y = f(x)$, what line, $\ell(x) = M(x-x_0) + f(x_0)$, through the point $(x_0, f(x_0))$ "best fits" it. In terms of magnitudes, we want to find $M$ such that
$$|f(x) - \ell(x)| =  |f(x) - f(x_0) - M(x-x_0)|$$
is appropriately "small" for all points $x$ that are close to $x_0$.
If we assume that $y=f(x)$ can be expressed as $$f(x) = a_0 + (x-x_0)a_1 + (x-x_0)^2 a_2 + (x-x_0)^3 a_3 + \cdots$$ then, if we can find a constant, $K$, such that
$$|f(x) - f(x_0) - M(x-x_0)| \le K(x-x_0)^2$$
it would be appropriate to define $f'(x_0) = M$. If we assume that $x \ne x_0$, then we can rewrite this as
$$\left|\dfrac{f(x) - f(x_0)}{x-x_0} - M\right| \le K|x-x_0|$$
for $x$ in some set of points that surrounds but doesn't include $x_0$.
Since we are only concerned about a neighborhood of $x_0$, we can assume that 
$K$ will exists as long as $(x-x_0)$ divides $\dfrac{f(x) - f(x_0)}{x-x_0} - M$
Examples
Let's take for example, $f(x)=x^2$.
$$\dfrac{f(x) - f(x_0)}{x-x_0} 
   = \dfrac{x^2 - x_0^2}{x-x_0}=x+x_0
   = 2x_0 + (x-x_0)$$
Hence 
$$\dfrac{f(x) - f(x_0)}{x-x_0} - 2x_0^2 = x-x_0$$
and we conclude $f'(x_0) = 2x_0$.

Next, let's take $f(x)=x^3$. Then
\begin{align}
   \dfrac{f(x) - f(x_0)}{x-x_0} 
   &= \dfrac{x^3 - x_0^3}{x-x_0} \\
   &= x^2 + x x_0 + x_0^2 \\
   &= 3x_0^2 + (x^2 + x x_0 - 2x_0^2) \\
   &= 3x_0^2 + (x-x_0)(x + 2x_0) \\
\end{align}
and we conclude $f'(x_0) = 3x_0^2$.
The obvious question is, "How did I get $3x_0^2$ out of $x^2 + x x_0 - 2x_0^2$ and, of course the answer is, evaluate $x^2 + x x_0 - 2x_0^2$ at $x=x_0$.

One more example would be $f(x) = \dfrac 1x$.
\begin{align}
   \dfrac{f(x) - f(x_0)}{x-x_0} 
   &= \dfrac{\dfrac 1x - \dfrac{1}{x_0}}{x-x_0} \\
   &= -\dfrac{x -x_0}{x x_0(x-x_0)} \\
   &= -\dfrac{1}{x x_0} \\
   &= -\dfrac{1}{x_0^2}+\dfrac{1}{x_0^2}-\dfrac{1}{x x_0} \\
   &= -\dfrac{1}{x_0^2}+\dfrac{1}{x x_0^2}(x-x_0) \\ \\
\end{align}
and we conclude that $f'(x_0) = -\dfrac{1}{x_0^2}$.

I'm not sure that this is a more reasonable alternative, but it is a viable alternative. And it has been worked out considerably by others. I'm pretty sure that there was an AMA article about it, but I don't have the resources to look it up.