---
tag: 5RY17F34YW
kind: proof
name: Proof of the power rule by induction
msc: [26A24]
proves: 0JZGW75C5Z
uses: [0F1WR9C83K, 0A61B60B5R, 0V9A43H7WV, 0GBZDAFPX8]
sources:
  - url: https://en.wikipedia.org/wiki/Power_rule
    locator: Proof by induction (natural numbers)
---

## Proof

We argue by induction on $n \ge 1$. Write $p_n(x) = x^n$, and use the convention $x^0 = 1$.

1. Base case $n = 1$. Fix $x \in \R$. For $h \ne 0$, the difference quotient is $\frac{(x + h) - x}{h} = 1$.
2. A function that is constantly $1$ near $0$ has limit $1$ at $0$. Indeed, in the [[epsilon-delta form of limits|0GBZDAFPX8]], any $\delta$ works since $\abs{1 - 1} = 0 < \varepsilon$.
3. So $p_1$ has [[derivative|0F1WR9C83K]] $1 = 1 \cdot x^0$ at $x$. Thus the claim holds for $n = 1$.
4. Induction step. Assume that $p_n$ is [[differentiable|0A61B60B5R]] at every $x$ with $p_n'(x) = n x^{n-1}$, for some $n \ge 1$.
5. For all $x$, $p_{n+1}(x) = x^n \cdot x = p_n(x) \, p_1(x)$.
6. By step 4, step 3 and the [[product rule|0V9A43H7WV]], $p_{n+1}$ is differentiable at every $x$, and
$$p_{n+1}'(x) = p_n'(x) \, p_1(x) + p_n(x) \, p_1'(x) = n x^{n-1} \cdot x + x^n \cdot 1.$$
7. The right side equals $n x^n + x^n = (n + 1) x^n$. This is the claim for $n + 1$.
8. By induction, $x \mapsto x^n$ is differentiable on $\R$ with derivative $n x^{n-1}$ for every $n \ge 1$.

## Idea

The case $n = 1$ is a direct computation. Writing $x^{n+1} = x^n \cdot x$, the product rule turns the formula for $n$ into the formula for $n + 1$.