Proof of the Power Rule

Electricity warning signIf you’ve got the word “power” in your name, you’d better believe expectations are going to be sky high for what you can do. The Power Rule in calculus brings it and then some.

The Power Rule, probably the most used rule when differentiating, gives us a drop dead simple way to differentiate polynomials. Specifically it says for that any polynomial term x raised to the power n with coefficient a:

(1)   \begin{equation*} \frac{d}{dx}ax^n = nax^{n-1}, \qquad n \neq 0 \end{equation*}

Apply this to every term in your polynomial, and you’ve got its derivative! Easy peasy. Let’s prove it.

We start with the definition of the derivative: the difference quotient.

(2)   \begin{equation*} \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \end{equation*}

Substituting in our polynomial:

    \[\lim_{\Delta x \to 0} \frac{a(x + \Delta x)^n - ax^n}{\Delta x}\]

For now, let’s just look at the numerator. Using the Binomial Theorem we can expand (x + \Delta x)^n. The numerator becomes:

    \[a[x^n + {n \choose 1}x^{n-1}\Delta x+ {n \choose 2}x^{n-2}(\Delta x)^2 + ... + {n \choose n-1}x(\Delta x)^{n-1} + (\Delta x)^n] - ax^n\]

Multiplying through by a and canceling the ax^n terms leaves:

    \[a{n \choose 1}x^{n-1}\Delta x +a{n \choose 2}x^{n-2}(\Delta x)^2 + ... + a{n \choose n-1}x(\Delta x)^{n-1} + a(\Delta x)^n\]

We can simplify this a little further since n choose 1 is equal to n. (I’m going to rearrange the order of a and n here, too, since we are coming into the homestretch, and I want to match the order I used when I presented the rule)

    \[nax^{n-1} \Delta x +{n \choose 2}ax^{n-2}(\Delta x)^2 ... +{n \choose n-1}a(\Delta x)^{n-1} + a(\Delta x)^n\]

Putting this back in as the numerator of our limit:

    \[\lim_{\Delta x \to 0} \frac{nax^{n-1} \Delta x +{n \choose 2}ax^{n-2}(\Delta x)^2 ... +{n \choose n-1}a(\Delta x)^{n-1} + a(\Delta x)^n}{\Delta x}\]

Divide by \Delta x:

    \[\lim_{\Delta x \to 0} nax^{n-1} +{n \choose 2}ax^{n-2}\Delta x + ... +{n \choose n-1}a(\Delta x)^{n-2} + a(\Delta x)^{n-1}\]

Suddenly we are left with a very simple limit. With the exception of the first term, every other term goes to 0 as \Delta x approaches zero. The first term, nax^{n-1},  is unaffected by any change in \Delta x. Thus, we have achieved our goal.

    \[\frac{d}{dx}ax^n = nax^{n-1}, \qquad n \neq 0 \]


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.