Proving The Positive Integer Power Rule

From a Difference Quotient to a General Rule

The identity function has derivative one, and constant functions have derivative zero. For powers higher than one, the derivative can also be found directly from the definition of a derivative. The key is to simplify the difference quotient before taking its limit.

For a function \(f\), its derivative at \(x\), when it exists, is the limit of the average rate of change over an increment \(h\) as that increment approaches zero:

For \(f(x)=x^n\), where \(n\) is a positive integer, the quotient initially contains a numerator that approaches zero and a denominator that approaches zero. Substituting \(h=0\) is therefore not the method. Instead, expand the power difference and factor out \(h\). After cancellation, the limit can be evaluated.

Worked Examples from the Definition

The following calculations illustrate the steps that will appear in the general proof: form the difference quotient, expand, factor out the increment, cancel it, and then take the limit.

The Factorization Behind the Rule

The examples suggest the general pattern, but a few examples cannot establish a rule for every positive integer exponent. The proof comes from a finite factorization that works for any such exponent.

To see why this identity holds, multiply \(a-b\) by the sum. The product with \(a\) gives the terms \(a^n,a^{n-1}b,\ldots,ab^{n-1}\). The product with \(-b\) gives \(-a^{n-1}b,-a^{n-2}b^2,\ldots,-b^n\). Every intermediate term cancels with an equal term of opposite sign, leaving \(a^n-b^n\).

For the derivative of \(x^n\), take \(a=x+h\) and \(b=x\). Then \(a-b=h\), so the identity factors the increment \(h\) out of the numerator.

Proof of the Positive Integer Power Rule

Fix a positive integer \(n\) and a real number \(x\). Begin with the difference quotient for \(f(x)=x^n\). For \(h\ne 0\), apply the difference-of-powers identity:

The cancellation is valid because the difference quotient is considered for \(h\ne 0\). The derivative is then found by taking the limit as \(h\to 0\). Each of the \(n\) terms in the remaining sum approaches \(x^{n-1}\). For example, the first term approaches \(x^{n-1}\), the next approaches \(x^{n-2}x=x^{n-1}\), and the last term is already \(x^{n-1}\). These limits follow from the limit laws for finite products and sums.

Since the sum has exactly \(n\) terms, its limit is the sum of \(n\) copies of \(x^{n-1}\):

This proves that the difference quotient has a limit for every real \(x\), and that the limit is \(nx^{n-1}\). Thus \(x^n\) is differentiable everywhere on the real line and its derivative is \(nx^{n-1}\).

Conditions and a Common Pitfall

The proof applies when \(n\) is a positive integer. That condition matters: it ensures the power difference expands into a finite sum with exactly \(n\) terms, which is what the argument uses. The result does not, by itself, establish a derivative rule for negative or non-integer exponents.

A common mistake is to substitute \(h=0\) into the original difference quotient and conclude that the derivative is \(0/0\), or that the derivative is undefined. For example, for \(x^2\) at \(x=3\), direct substitution gives

This does not evaluate the limit. It only shows why substitution must wait until after simplification. The correct calculation factors the numerator:

and then takes the limit, which equals \(6\). Cancellation is performed for nonzero \(h\); the simplified expression can then be used to find the limit as \(h\) approaches zero.

Another tempting error is to assume \((x+h)^n=x^n+h^n\). This is false in general: when \(n=2\), \((x+h)^2=x^2+2xh+h^2\), and the middle term \(2xh\) cannot be omitted. The correct expansion or difference-of-powers factorization keeps all terms, allowing the factor \(h\) to be identified accurately.

Why the Derivation Is Useful

The power rule is more than a convenient formula. Its derivation explains exactly how the change in \(x^n\) depends on a small change in \(x\): after the increment \(h\) is factored out, the quotient approaches \(n\) copies of \(x^{n-1}\). The factor \(n\) comes from the number of terms in the factorization, while the remaining power \(x^{n-1}\) is the limit of each term.

This reasoning also gives a reliable way to check the rule. For a particular positive integer exponent, one can return to the difference quotient, simplify it without setting \(h\) equal to zero, and confirm that its limit matches \(nx^{n-1}\). The method is especially valuable when first learning differentiation because it connects the general formula to the defining limit rather than treating the formula as an unexplained instruction.