Deriving the Positive-Integer Power Rule

Why Derive the Power Rule?

A differentiation rule is more useful when you know where it comes from. The positive-integer power rule can be obtained directly from the definition of the derivative: expand \((x+h)^n\), subtract \(x^n\), divide by \(h\), and take a limit.

The binomial theorem organizes the expansion so that the cancellation and the limit are easy to see. The key is to divide by \(h\) before letting \(h\) approach zero. Once that is done, the first surviving term gives the derivative, while every other term contains a positive power of \(h\) and tends to zero.

The Difference Quotient and the Binomial Theorem

For a function \(f\), its derivative at \(x\), when the limit exists, is defined by

Here \(h\) is a nonzero change in the input while the quotient is formed. The limit then describes what happens as that change approaches zero. For this derivation, take \(f(x)=x^n\), where \(n\) is a positive integer.

The binomial theorem states that for a positive integer \(n\),

In this formula, \(\binom{n}{k}\) is the binomial coefficient, and the terms at the beginning of the expansion are

The term \(x^n\) is the term with \(k=0\). It will cancel when we subtract \(f(x)=x^n\). The next term, \(nx^{n-1}h\), will become \(nx^{n-1}\) after division by \(h\). The later terms still contain powers of \(h\).

Worked Example: Deriving the Rule for \(x^2\)

Worked Example: Deriving the Rule for \(x^3\)

Worked Example: Deriving the Rule for \(x^4\)

Derivation for Every Positive Integer

The examples show the same structure for several exponents. The binomial theorem lets us establish it for an arbitrary positive integer \(n\), rather than relying on a pattern from a few examples.

In equations, start with

The \(k=1\) term is \(\binom{n}{1}x^{n-1}h^0=nx^{n-1}\). For every \(k\) from \(2\) through \(n\), the exponent \(k-1\) is positive. Since \(x\) and the binomial coefficients are fixed while \(h\) approaches zero, each such term tends to zero. There are only finitely many terms, so the limit of their sum is the sum of their limits. Therefore,

This proof also covers \(n=1\). In that case the sum has only the \(k=1\) term, and the difference quotient is \(1\), which agrees with \(1\cdot x^0=1\). No additional terms need to vanish.

What the Derivation Requires

The argument depends on \(n\) being a positive integer. That condition allows \((x+h)^n\) to be expanded as a finite binomial sum with powers \(h^0,h^1,\ldots,h^n\). After division by \(h\), the first term gives the claimed derivative and each later term has a positive power of \(h\).

The derivation works for every real \(x\): no step divides by \(x\), and every expression in the finite expansion is defined for real \(x\). The only division by a variable increment is by \(h\), which is allowed in the difference quotient because \(h\ne 0\); the limit is taken afterward.

This proof does not establish a power rule for zero, negative, or fractional exponents. Those cases require other arguments and conditions. The conclusion here is precisely the positive-integer rule.

A Common Mistake: Taking Limits Before Dividing

For \(f(x)=x^3\), someone might observe that \((x+h)^3-x^3\) approaches zero as \(h\) approaches zero and conclude that the derivative is zero. That is a concrete but incorrect calculation:

The derivative is not the limit of the numerator alone. It is the limit of the ratio of the numerator to \(h\). The numerator and denominator both approach zero, so their separate limits do not determine the limit of the quotient. Divide first:

The cancellation of one factor of \(h\) matters: the term \(3x^2h\) in the numerator becomes \(3x^2\), which does not disappear in the limit. This is why the order of the steps is essential.

Why the Power Rule Matters

The derivation connects a general differentiation formula to the definition of a derivative. Rather than memorizing the coefficient \(n\) and the exponent \(n-1\) as an unexplained pattern, you can identify where each comes from: \(n\) is the coefficient of the term linear in \(h\), and the exponent drops by one when that term is divided by \(h\).

The proof also illustrates a general strategy for limits of difference quotients: simplify the quotient while the increment is nonzero, then take the limit. For this topic, the binomial theorem produces a finite expression, so the terms that vanish are visible individually.

The result is a rule for differentiating \(x^n\) when \(n\) is a positive integer. Applying it as part of longer expressions is a separate task; the essential result here is its derivation from the derivative’s definition.