On-line Math 21

2.2  Differentiation Rules

Theorem 2 [The Power Rule]. For each positive integer n , (xⁿ)¢ = nx^n-1.

Proof:

The hint here is to look at the binomial expansion,

(a+b)n = an+nan-1b+ n(n-1) 2 an-2b2+¼.

Where did that come from? If you recall Pascal's triangle,

0,  1 1,  1 1 2,  1 2 1 3,  1 3 3 1 4,  1 4 6 4 1

which is supposed to give the coefficients of (a+b)ⁿ where n is the number on the left, then the coefficients written out here are correct. You should probably check this for, say, n = 3 by multiplying it out. The formula I used for the n^th power is the general expression for what shows up in the first three terms of Pascals triangle in each row.

Then, apply that to the proof, with (x+h) instead of (a+b) :

For f(x) = xⁿ ,

f¢(x) = lim h® 0  f(x+h)-f(x) h = lim h® 0  (x+h)n-(x)n h = lim h® 0  xn+n xn-1h+... +hn-xn h ,

You then cancel the xⁿ 's and it simplifies easily.

Copyright (c) 2000 by David L. Johnson.