Proof of the formula for the derivative of a polynomial

Proof

We argue by induction on $n$, the largest index of the coefficients.

1. Base case $n = 0$. Then $p(x) = a_0$ is constant. For $h \ne 0$, its difference quotient at $x$ is $\frac{a_0 - a_0}{h} = 0$.
2. A function that is constantly $0$ near $0$ has limit $0$ at $0$, by the epsilon-delta form of limits. So $p$ has derivative $0$ at every $x$.
3. The claimed formula for $n = 0$ is the empty sum $\sum_{k=1}^{0} k a_k x^{k-1} = 0$. So the base case holds.
4. Induction step. Assume the claim for all polynomials $\sum_{k=0}^{n} a_k x^k$, for some $n \ge 0$. Let $p(x) = \sum_{k=0}^{n+1} a_k x^k$.
5. Write $p = r + a_{n+1} s$, where $r(x) = \sum_{k=0}^{n} a_k x^k$ and $s(x) = x^{n+1}$.
6. By the induction hypothesis, $r$ is differentiable on $\R$ with $r'(x) = \sum_{k=1}^{n} k a_k x^{k-1}$.
7. By the power rule with exponent $n + 1 \ge 1$, $s$ is differentiable on $\R$ with $s'(x) = (n + 1) x^{n}$.
8. By the linearity of differentiation with coefficients $1$ and $a_{n+1}$, $p$ is differentiable at every $x$, and $$p'(x) = \sum_{k=1}^{n} k a_k x^{k-1} + (n + 1) a_{n+1} x^{n} = \sum_{k=1}^{n+1} k a_k x^{k-1}.$$
9. This is the claim for $n + 1$. By induction, it holds for every $n$.