Algebraic proof of the chain rule? I would like to prove the chain rule: given $f$ and $g$ polynomial functions, $h = f \circ g$, and $a \in \mathbb{R}$, that $h'(a) = f'(g(a)) \cdot g'(a)$.  However, I would like to do so without using the limit definition of the derivative or any sort of differentiation rules.
So far, the only lead I've got is that given $P(x)$ a polynomial function, by the division algorithm, $P(x) = (x-a)^2Q(x) + R(x)$, and $R(x)$ is the equation of the line tangent to $P(x)$ at $x = a$.

This is not a full solution, but it does describe another approach to defining the derivative of a polynomial that does not involve limits, and that is related the division algorithm.
Let $P(x)$ be any polynomial.  Choose some real number $a$.  Then if we divide $P(x)$ by $x-a$ we get a quotient, $Q(x)$, and a constant remainder, $R$.  In fact by the Remainder Theorem $R=P(a)$, so we have $P(x) = (x-a)Q(x) + P(a)$.  Rearranging,
$$Q(x) = \frac{P(x)-P(a)}{x-a}$$
This equation has a natural interpretation:  the quotient polynomial $Q(x)$ tells us the slope of the secant line passing through the graph of $P(x)$ at the points $(a,P(a))$ and $(x,P(x))$.  With this interpretation, we can recognize that the slope of the tangent line to $P(x)$ at $x=a$ is just given by $Q(a)$.  So we define $f'(a) = Q(a)$.  (Since $Q(x)$ is a polynomial, there is no need to take a limit here.)
With this as background, let's set out to answer the question in the OP.
To compute the derivative of $f(g(x))$ at $x=a$ we divide $f(g(x))$ by $x-a$, obtaining a quotient $q_1(x)$, with
$$f(g(x))=(x-a)q_1(x) + f(g(a))$$
and then the derivative is $q_1(a)$.
To compute $f'(g(a))$ we divide $f(x)$ by $x-g(a)$, obtaining a quotient $q_2(x)$, with $$f(x) = (x-g(a))q_2(x) + f(g(a))$$
and then $f'(g(a)) = q_2(g(a))$.
To compute $g'(a)$ we divide $g(x)$ by $x-a$, obtaining a quotient $q_3(x)$, with $$g(x) = (x-a)q_3(x) + g(a)$$
and then $g'(a) = q_3(a)$.
The chain rule is then expressed by the identity $$q_1(a) = q_2(g(a))\cdot q_3(a)$$.  This is what we need to prove.  Can you take it from here?
