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  "body": "\n## Proof\n\nWe argue by induction on $n$, the largest index of the coefficients.\n\n1. Base case $n = 0$. Then $p(x) = a_0$ is constant. For $h \\ne 0$, its difference quotient at $x$ is $\\frac{a_0 - a_0}{h} = 0$.\n2. A function that is constantly $0$ near $0$ has limit $0$ at $0$, by the [[epsilon-delta form of limits|0GBZDAFPX8]]. So $p$ has [[derivative|0F1WR9C83K]] $0$ at every $x$.\n3. 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.\n4. Induction step. Assume the claim for all polynomials $\\sum_{k=0}^{n} a_k x^k$, for some $n \\ge 0$. Let $p(x) = \\sum_{k=0}^{n+1} a_k x^k$.\n5. Write $p = r + a_{n+1} s$, where $r(x) = \\sum_{k=0}^{n} a_k x^k$ and $s(x) = x^{n+1}$.\n6. By the induction hypothesis, $r$ is [[differentiable|0A61B60B5R]] on $\\R$ with $r'(x) = \\sum_{k=1}^{n} k a_k x^{k-1}$.\n7. By the [[power rule|0JZGW75C5Z]] with exponent $n + 1 \\ge 1$, $s$ is differentiable on $\\R$ with $s'(x) = (n + 1) x^{n}$.\n8. By the [[linearity of differentiation|0M0TKFRX2A]] with coefficients $1$ and $a_{n+1}$, $p$ is differentiable at every $x$, and\n$$p'(x) = \\sum_{k=1}^{n} k a_k x^{k-1} + (n + 1) a_{n+1} x^{n} = \\sum_{k=1}^{n+1} k a_k x^{k-1}.$$\n9. This is the claim for $n + 1$. By induction, it holds for every $n$.\n\n## Idea\n\nA polynomial is built by adding one term $a_k x^k$ at a time. The power rule differentiates each term, and linearity lets us differentiate the sum term by term.\n",
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