# proof of the power rule

The power rule^{} can be derived by repeated application of the product rule^{}.

## Proof for all positive integers $n$

The power rule has been shown to hold for $n=0$ and $n=1$. If the power rule is known to hold for some $k>0$, then we have

$\frac{\mathrm{d}}{\mathrm{d}x}}{x}^{k+1$ | $=$ | $\frac{\mathrm{d}}{\mathrm{d}x}}(x\cdot {x}^{k})$ | ||

$=$ | $x\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}{x}^{k}\right)+{x}^{k}$ | |||

$=$ | $x\cdot (k{x}^{k-1})+{x}^{k}$ | |||

$=$ | $k{x}^{k}+{x}^{k}$ | |||

$=$ | $(k+1){x}^{k}$ |

Thus the power rule holds for all positive integers $n$.

## Proof for all positive rationals $n$

Let $y={x}^{p/q}$. We need to show

$$\frac{\mathrm{d}y}{\mathrm{d}x}({x}^{p/q})=\frac{p}{q}{x}^{p/q-1}$$ | (1) |

The proof of this comes from implicit differentiation^{}.

By definition, we have ${y}^{q}={x}^{p}$. We now take the derivative with respect to $x$ on both sides of the equality.

$\frac{\mathrm{d}}{\mathrm{d}x}}{y}^{q$ | $=$ | $\frac{\mathrm{d}}{\mathrm{d}x}}{x}^{p$ | ||

$\frac{\mathrm{d}}{\mathrm{d}y}}({y}^{q}){\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ | $=$ | $p{x}^{p-1}$ | ||

$q{y}^{q-1}{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$ | $=$ | $p{x}^{p-1}$ | ||

$\frac{\mathrm{d}y}{\mathrm{d}x}$ | $=$ | $\frac{p}{q}}{\displaystyle \frac{{x}^{p-1}}{{y}^{q-1}}$ | ||

$=$ | $\frac{p}{q}}{x}^{p-1}{y}^{1-q$ | |||

$=$ | $\frac{p}{q}}{x}^{p-1}{x}^{p(1-q)/q$ | |||

$=$ | $\frac{p}{q}}{x}^{p-1+p/q-p$ | |||

$=$ | $\frac{p}{q}}{x}^{p/q-1$ |

## Proof for all positive irrationals $n$

For positive irrationals we claim continuity due to the fact that (1) holds for all positive rationals, and there are positive rationals that approach any positive irrational.

## Proof for negative powers $n$

We again employ implicit differentiation. Let $u=x$, and differentiate ${u}^{n}$ with respect to $x$ for some non-negative $n$. We must show

$$\frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}=-n{u}^{-n-1}$$ | (2) |

By definition we have ${u}^{n}{u}^{-n}=1$. We begin by taking the derivative with respect to $x$ on both sides of the equality. By application of the product rule we get

$\frac{\mathrm{d}}{\mathrm{d}x}}({u}^{n}{u}^{-n})$ | $=$ | $1$ | ||

${u}^{n}{\displaystyle \frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}}+{u}^{-n}{\displaystyle \frac{\mathrm{d}{u}^{n}}{\mathrm{d}x}}$ | $=$ | $0$ | ||

${u}^{n}{\displaystyle \frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}}+{u}^{-n}(n{u}^{n-1})$ | $=$ | $0$ | ||

${u}^{n}{\displaystyle \frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}}$ | $=$ | $-n{u}^{-1}$ | ||

$\frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}$ | $=$ | $-n{u}^{-n-1}$ |

Title | proof of the power rule |
---|---|

Canonical name | ProofOfThePowerRule |

Date of creation | 2013-03-22 12:28:06 |

Last modified on | 2013-03-22 12:28:06 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 9 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 26A24 |

Related topic | ProductRule |

Related topic | Derivative2 |