## The Power Rule

 The Power Rule If a is any real number, and f(x) = xa, then f'(x) = axa-1.

The proof is divided into several steps. However, you can skip to the last step for a quick proof that uses the formula for the derivative of exponential functions.

Step 1: Proof of the Power Rule for Non-Negative Integer Exponents

In this step, we assume that f(x) = xn, where n is some positive integer: 0, 1, 2, 3, ....

First, if n happens to be zero, f(x) = x0 = 1, a constant, and so its derivative is zero, by the result we proved in the text.

Thus, assume that n is a positive integer. To follow the proof of this step you must recognize a nice little algebraic fact. First look at these identities.

a1 - b1 = (a-b)
a2 - b2 = (a-b)(a+b)
a3 - b3 = (a-b)(a2 + ab + b2)
a4 - b4 = (a-b)(a3 + a2b + ab2 + b2)
. . . etc.

(Use the distributive law to expand the right-hand side in each case.)

These examples generalize to give us the following formula

 Difference of Two n th Powers If a and b are real numbers, and n is a positive integer, then an - bn = (a-b)(an-1 + an-2b + an-3b2 + . . . + abn-2 + bn-1).

Now let us turn back to the proof at hand. Write f(x) = xn. Then

f'(x) =  limh0 f(x+h) - f(x)h
=  limh0 (x+h)n - xnh)

We now rewrite the numerator using the identity above with "a" replaced by the quantity (x+h) and "b" by x, getting

=  limh0 [(x+h) - x][(x+h)n-1 + (x+h)n-2x + . . . + xn-1]h
=  limh0 h[(x+h)n-1 + (x+h)n-2x + . . . + xn-1]h
Since (x+h)-x = h
=  limh0 [(x+h)n-1 + (x+h)n-2x + . . . + xn-1]
Cancel the h

Now that we have canceled the h from the denominator, we have a closed-form function, so we can evaluate the limit by substituting h = 0. Each term in the sum becomes xn-1 if we do that, so we get

f'(x) = xn-1 + xn-1 + . . . + xn-1.

There are n terms, so f'(x) = nxn-1, completing the proof of this step.

Step 2: Proof of the Power Rule for Negative Integer Exponents

Here, we assume that f(x) = x-n, where n = 0, 1, 2, 3, ... . Since we are required here to justify the power rule for negative integers, we can't simply go ahead and use it! "Officially," all we can use are the power rule for positive integers (Step 1), and the product and quotient rules. (The product rule is proved in the text, and the quotient rule is proved here.)

What we can do is this: write x-n as 1/xn and use the quotient rule. Applying the quotient rule to 1/xn gives

d

dx
1

xn
=  (0)(xn) - (1)(nxn-1)(xn)2
=  -nxn-1x2n
=  -nxn-1-2n = -nx-n-1,

and we are done with this step.

Step 3: Proof of the Power Rule for Rational Exponents

For this step, we assume that f(x) = xp/q, where p and q are integers (positive or negative), and we must show that

 f'(x) = pq xp/q - 1 .

To prove this, let y = xp/q. Thus, the problem is to calculate dy/dx without assuming the power rule for anything but integer exponents. Before we do anything, we raise both sides to the power q in order to get integer exponents everywhere.

yq = (xp/q)q = xp

Now we take the equation yq = xp and differentiate both sides with respect to x. By the chain rule,

 ddx (yq) = qyq-1 dydx ,

whereas

 ddx (xp) = pxp-1 .

Equating these derivatives gives

 qyq-1 dydx = pxp-1.

Now remember that we want dy/dx by itself. We can solve for dy/dx by dividing both sides by the quantity qyq-1:

 dydx = pxp-1qyq-1 .

But y = xp/q, and so

d

dx
(xp/q) =  pxp-1q(xp/q)q-1
=  pxp-1qxp - p/q
=  pq xp-1-(p-p/q)
=  pq xp/q - 1.
Done!

Step 4: Proof of the Power Rule for Arbitrary Real Exponents (The General Case)

Actually, this step does not even require the previous steps, although it does rely on the use of exponential functions and their derivatives.

First, we need the equality xn = enlnx. (You can check this by taking the natural logarithm of both sides.)

d

dx
xn =  ddx enlnx
=  enlnx ddx [n lnx]
By the derivative rule for exponential functions
=  enlnx nx
By the derivative rule for logarithmic functions
=  xn nx
By the equality xn = enlnx
=  nxn-1.

This proves the power rule for all real powers.