proof of the power rule

The power ruleMathworldPlanetmathPlanetmath can be derived by repeated application of the product ruleMathworldPlanetmath.

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

ddxxk+1 = ddx(xxk)
= x(ddxxk)+xk
= x(kxk-1)+xk
= kxk+xk
= (k+1)xk

Thus the power rule holds for all positive integers n.

Proof for all positive rationals n

Let y=xp/q. We need to show

dydx(xp/q)=pqxp/q-1 (1)

The proof of this comes from implicit differentiationMathworldPlanetmath.

By definition, we have yq=xp. We now take the derivative with respect to x on both sides of the equality.

ddxyq = ddxxp
ddy(yq)dydx = pxp-1
qyq-1dydx = pxp-1
dydx = pqxp-1yq-1
= pqxp-1y1-q
= pqxp-1xp(1-q)/q
= pqxp-1+p/q-p
= pqxp/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 un with respect to x for some non-negative n. We must show

du-ndx=-nu-n-1 (2)

By definition we have unu-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

ddx(unu-n) = 1
undu-ndx+u-ndundx = 0
undu-ndx+u-n(nun-1) = 0
undu-ndx = -nu-1
du-ndx = -nu-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