Power Rule

From the Identity to Higher Powers

In “Derivative of the Identity,” the difference quotient simplified exactly to one. For a higher power, the quotient does not usually stay constant, but its algebraic structure can still be exposed before taking a limit. The key is to factor a difference of powers. This gives the power rule for integer exponents directly from the definition of the derivative.

We begin with nonnegative integer powers. For a positive integer \(n\), the function \(x\mapsto x^n\) is defined on all of \(\mathbb{R}\), so its derivative can be considered at every real input. Negative integer powers are also covered, but their domains exclude zero. The rule for arbitrary real exponents is not established here; it requires additional results about how such powers are defined and how they vary.

The algebraic identity we need is the difference-of-powers factorization. If \(u\) and \(v\) are real numbers and \(n\) is a positive integer, then

To check the identity, multiply out the right-hand side. The terms \(u^{n-j}v^j\) from the first product cancel with the matching terms \(-u^{n-1-j}v^{j+1}\) from the second. Only \(u^n\) and \(-v^n\) remain. This factorization turns the difference quotient into a finite sum whose limit is straightforward to evaluate.

The Power Rule for Nonnegative Integer Exponents

Proof. If \(n=0\), \(p_0\) is the constant function with value \(1\). By the Derivative of a Constant Function Theorem, its derivative is \(0\) at every interior point of its domain, and every real number is an interior point of \(\mathbb{R}\).

Now let \(n\geq1\) and fix \(a\in\mathbb{R}\). For every nonzero \(h\), apply the difference-of-powers identity with \(u=a+h\) and \(v=a\). Since \(u-v=h\), the difference quotient is

As \(h\to0\), each term in this finite sum tends to \(a^{n-1-j}a^j=a^{n-1}\). There are exactly \(n\) terms, so the limit of the sum is \(n a^{n-1}\). Thus the derivative exists and equals \(n a^{n-1}\). Since \(a\) was arbitrary, the result holds at every real input. \(\square\)

The proof includes the point \(a=0\); it does not require dividing by \(a\). This matters because factoring out a power of \(a\) in the quotient could create a division by zero and leave the derivative at zero unproved. The finite-sum argument works there just as it does elsewhere. When \(n=1\), the formula gives \(1\), in agreement with the Derivative of the Identity Theorem.

Negative Integer Powers

For a negative exponent, zero must be excluded from the domain: \(x^{-m}=1/x^m\) is undefined when \(x=0\). At any nonzero point, however, the same difference-of-powers factorization yields the derivative. We give the argument directly from the difference quotient, so no quotient rule is needed.

Proof. Fix \(a\ne0\). If \(|h|<|a|/2\), then \(|a+h|\geq |a|-|h|>|a|/2>0\), so \(a+h\) remains in the domain. For such \(h\ne0\), rewrite and factor the difference quotient:

In the last equality we used the difference-of-powers identity. As \(h\to0\), each of the \(m\) terms in the numerator sum tends to \(a^{m-1}\), so the numerator tends to \(m a^{m-1}\). The denominator tends to \(a^m a^m=a^{2m}\), which is nonzero because \(a\ne0\). Therefore the quotient has limit

This is the derivative at \(a\), and \(a\) was any nonzero real number. \(\square\)

Using the Rule Carefully

For integer exponents, the two results can be summarized as \( \frac{d}{dx}x^n=n x^{n-1}\), with the domain understood. If \(n\) is positive, \(x^n\) is defined for every real \(x\) and the formula holds everywhere. If \(n=0\), the function is constant and its derivative is \(0\); the expression \(0x^{-1}\) should not be used at zero, where \(x^{-1}\) is undefined. If \(n\) is negative, both the function and its derivative formula are restricted to nonzero inputs.

A common pitfall is to treat the displayed formula as if it automatically applied to every exponent and every input. The proofs here use integer powers and, for negative powers, a denominator that stays nonzero near the point under consideration. They do not prove a differentiation rule for fractional or irrational exponents. Nor does the notation \(x^{n-1}\) remove a domain restriction: for negative powers, zero remains excluded.

The power rule is useful because it converts a limit problem into a short calculation once its hypotheses are checked. The difference quotient supplies the justification; the formula is not an independent shortcut detached from the definition. Later differentiation rules will let us combine power functions with other expressions, but this result is the basic calculation on which those methods build.