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 $x = a$.
 A: Recall that the derivative is linear and satisfies
$$(fg)'=f'g+fg'$$
Using this coupled with induction you can show that the derivative of $g(x)^n$ is
$$ng(x)^{n-1}g'(x)$$
Thus if
$$f(x)=\sum{a_nx^n}$$
then
$$(f(g(x)))'=\sum{na_ng(x)^{n-1}g'(x)}=f'(g(x))g'(x)$$
This is precisely the chain rule.
A: Let $D:k[X]\to k[X]$ be the usual derivative, which is the unique $k$-linear map such that $D(fg)=D(f)g+fD(g)$ for all $f$, $g\in k[X]$ and $D(X)=1$. We want to show that $$D(f\circ g)=D(f)\circ g \cdot D(g)\qquad\text{for all $f$, $g\in k[X]$.}$$ Since $D$ is $k$-linear and composition is $k$-linear and every $f\in k[X]$ is a linear combination of monomials, it is enough that we prove this when $f$ is $X^n$, that is (since we know that $D(X^n)=nX^{n-1}$) that $$D( g^n)=n g^{n-1} \cdot D(g)\qquad\text{for all $g\in k[X]$ and all $n\in\mathbb N_0$.}$$
This you can easily prove by induction on $n$. Indeed, i is obvious that it holds if $n=0$, and if it holds for some $n$ we have that 
\begin{align}
D(g^{n+1})
&= D(g^n\cdot g)\\
&= D(g^n)\cdot g + g^n D(g) \\
&= ng^{n-1}D(g)\cdot g+g^n D(g) \\
&= (n+1)g^nD(g).
\end{align} 
A: With differential of a function $y=f(u)$, 
$$dy=f'(u)du$$
and also for $u=g(x)$
$$du=g'(x)dx$$
so with substituation 
$$dy=f'(g(x))g'(x)dx$$
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?
