---
tag: R315P11FTH
kind: proof
name: Proof of the derivative of x^a by the chain rule
msc: [26A24, 26A09]
proves: 0KY08TWDMQ
uses: [0TV07AQT62, 0RXX2PT1MQ, 0EX9GFJ0B3, 07X1QA6BRW, 0Z9JB71BS0, 0E8HNVBCSP, 0RKW2X63YR, 0RMVS7RXZX]
sources:
  - url: https://en.wikipedia.org/wiki/Power_rule
    locator: Proof for real exponents
---

## Proof

1. Fix a real number $a$. For $x > 0$, the [[real power|0TV07AQT62]] is $x^a = e^{a \ln x}$.
2. Let $u(x) = a \ln x$ on the open interval $(0, \infty)$. By the [[derivative of the logarithm|07X1QA6BRW]] and the [[constant multiple rule|0Z9JB71BS0]], $u$ is differentiable with $u'(x) = \frac{a}{x}$.
3. By [[$\exp' = \exp$|0EX9GFJ0B3]], $\exp$ is differentiable on $\mathbb{R}$ with derivative $\exp$.
4. By the [[chain rule|0RXX2PT1MQ]], $x \mapsto x^a = \exp(u(x))$ is differentiable at every $x > 0$, with derivative $e^{a \ln x} \cdot \frac{a}{x} = \frac{a \, x^a}{x}$.
5. It remains to show $\frac{x^a}{x} = x^{a-1}$. By the [[exponential of a sum|0E8HNVBCSP]], $e^{\ln x} e^{-\ln x} = e^{0} = 1$, where $e^0 = 1$ by the [[series of the exponential|0RMVS7RXZX]].
6. Since [[$e^{\ln x} = x$|0RKW2X63YR]], this gives $e^{-\ln x} = \frac{1}{x}$.
7. Again by the exponential of a sum, $x^{a-1} = e^{a \ln x + (-\ln x)} = e^{a \ln x} e^{-\ln x} = \frac{x^a}{x}$.
8. By steps 4 and 7, $\frac{d}{dx} x^a = a x^{a-1}$ for every $x > 0$.

## Idea

Rewrite the power as an exponential, $x^a = e^{a \ln x}$, so that the exponent becomes an ordinary function of $x$. The chain rule then differentiates it, and the factor $\frac{1}{x}$ from $\ln$ lowers the power by one.