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The Derivative from First Principles · Tutorial 345 of 1495

The Integer-Power Derivative from the Binomial Theorem

Learn how the binomial expansion turns the difference quotient for an integer power into a limit whose surviving term gives its derivative.

Intermediate 9 min read

What You'll Learn

  • State the derivative definition used for the function \(f(x)=x^n\).
  • Apply the binomial theorem when the exponent is a positive integer.
  • Cancel the constant term and simplify the difference quotient.
  • Explain why every term beyond the first tends to zero.
  • Compute and check derivatives at positive, negative, and zero inputs.
  • Identify the error in taking the limit before dividing by the increment.

Why Expand an Integer Power?

The derivative of a function at a point describes the limiting slope of its secant lines there. For a function given by an integer power, the difference quotient initially contains a power of \(x+h\). Expanding that power reveals exactly how the quotient behaves as the increment \(h\) approaches zero.

The aim here is to derive the formula for the derivative of \(f(x)=x^n\), where \(n\) is any positive integer. The exponent must be a positive integer for the binomial expansion used below to consist of a finite sum. The derivation will apply to every real input \(x\), including \(x=0\).

Definition: For a function \(f\) defined near \(x\), its derivative at \(x\), when the limit exists, is $$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. $$ The quotient is considered for \(h\ne 0\); the derivative is the limit as such nonzero increments approach zero.

For \(f(x)=x^n\), substituting into this definition gives

$$ f'(x)=\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}, \qquad h\ne 0. $$

The numerator is zero when \(h=0\), so directly substituting \(h=0\) into the quotient does not evaluate the limit. Instead, we expand the numerator and simplify while \(h\) is still nonzero.

The Binomial Expansion

For a positive integer \(n\), the binomial theorem states that

$$ (x+h)^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}h^k. $$

Here \(k\) takes the integer values from \(0\) to \(n\), and \(\binom{n}{k}\) is the binomial coefficient. In particular, the first two terms are \(x^n\) and \(nx^{n-1}h\). The remaining terms, if there are any, contain \(h^2,h^3,\ldots,h^n\).

The \(k=0\) term is exactly \(x^n\), which is the term subtracted in the difference quotient. Removing it leaves only terms that contain at least one factor of \(h\). Since the quotient is for \(h\ne0\), we can divide each remaining term by \(h\).

Derivation of the Integer-Power Derivative

Apply the binomial theorem to the difference quotient. The term with \(k=0\) cancels with the subtracted \(x^n\). For every remaining term, dividing by \(h\) lowers its exponent of \(h\) by one:

$$ \begin{aligned} \frac{(x+h)^n-x^n}{h} &= \frac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}x^{n-k}h^k-x^n}{h}\\ &= \frac{\displaystyle\sum_{k=1}^{n}\binom{n}{k}x^{n-k}h^k}{h}\\ &= \sum_{k=1}^{n}\binom{n}{k}x^{n-k}h^{k-1}. \end{aligned} $$

The \(k=1\) term is \(\binom{n}{1}x^{n-1}h^0=nx^{n-1}\). Every term with \(k\ge2\) contains the factor \(h^{k-1}\), a positive power of \(h\). For fixed real \(x\), those terms tend to zero as \(h\to0\). There are only finitely many of them, so their sum tends to zero as well.

Theorem: If \(n\) is a positive integer and \(f(x)=x^n\) for real \(x\), then \(f\) is differentiable at every real \(x\), and $$ f'(x)=nx^{n-1}. $$

To see the limit argument explicitly, separate the first term from the others:

$$ \begin{aligned} f'(x) &=\lim_{h\to0}\sum_{k=1}^{n}\binom{n}{k}x^{n-k}h^{k-1}\\ &=nx^{n-1} +\lim_{h\to0}\sum_{k=2}^{n}\binom{n}{k}x^{n-k}h^{k-1}\\ &=nx^{n-1}. \end{aligned} $$

If \(n=1\), the sum from \(k=2\) to \(n\) has no terms, and the quotient is simply \(1\). The formula then reads \(1\cdot x^0=1\), using \(x^0=1\) in this expression, including at \(x=0\). If \(n>1\) and \(x=0\), the same expansion remains valid: the only potentially nonzero quotient term is \(h^{n-1}\), which tends to zero. Thus the result also covers the origin without needing to divide by \(x\).

Worked Examples

Worked Example: The Derivative of \(x^4\) at \(x=2\)

Let \(f(x)=x^4\). At \(x=2\), the difference quotient is

$$ \frac{f(2+h)-f(2)}{h} = \frac{(2+h)^4-2^4}{h}. $$

Expand \((2+h)^4\) using the binomial theorem:

$$ (2+h)^4 = 2^4+4(2^3)h+6(2^2)h^2+4(2)h^3+h^4 = 16+32h+24h^2+8h^3+h^4. $$

Since \(2^4=16\), subtract and divide by \(h\ne0\):

$$ \begin{aligned} \frac{(2+h)^4-16}{h} &=\frac{32h+24h^2+8h^3+h^4}{h}\\ &=32+24h+8h^2+h^3. \end{aligned} $$

Now take the limit:

$$ f'(2)=\lim_{h\to0}(32+24h+8h^2+h^3)=32. $$

Check against the derived formula: \(4(2)^{4-1}=4\cdot8=32\). This agrees with the difference-quotient limit.

Worked Example: A Negative Input for \(x^3\)

Let \(f(x)=x^3\) and evaluate its derivative at \(x=-1\). Begin with

$$ \frac{f(-1+h)-f(-1)}{h} = \frac{(-1+h)^3-(-1)^3}{h}. $$

Expanding gives

$$ (-1+h)^3=(-1)^3+3(-1)^2h+3(-1)h^2+h^3 =-1+3h-3h^2+h^3. $$

Because \((-1)^3=-1\), the constant terms cancel. For \(h\ne0\),

$$ \begin{aligned} \frac{(-1+h)^3-(-1)}{h} &=\frac{3h-3h^2+h^3}{h}\\ &=3-3h+h^2. \end{aligned} $$

Therefore,

$$ f'(-1)=\lim_{h\to0}(3-3h+h^2)=3. $$

The formula gives \(3(-1)^{3-1}=3(-1)^2=3\), confirming the value obtained directly from the quotient.

Worked Example: The Derivative of \(x^5\) at the Origin

Let \(f(x)=x^5\) and take \(x=0\). The difference quotient becomes

$$ \frac{f(0+h)-f(0)}{h} = \frac{h^5-0^5}{h}. $$

For \(h\ne0\), cancel one factor of \(h\):

$$ \frac{h^5}{h}=h^4. $$

Taking the limit gives

$$ f'(0)=\lim_{h\to0}h^4=0. $$

The general formula gives \(5(0)^{5-1}=5\cdot0^4=0\). This agrees with the limit. Notice that the cancellation used \(h\ne0\), not \(x\ne0\), so evaluating at the origin causes no difficulty.

Why the Argument Works

The expansion does more than produce a convenient algebraic form. It isolates the part of the change in \(x^n\) that is proportional to the small increment. The first-order term is \(nx^{n-1}h\). The other terms involve at least \(h^2\); after dividing by \(h\), they still contain a positive power of \(h\), so they disappear in the limit.

In the difference quotient, dividing by \(h\) measures change per unit of input change. The binomial expansion shows that this ratio approaches the coefficient of the linear-in-\(h\) term. That coefficient, \(nx^{n-1}\), is therefore the limiting slope. This explains why the exponent becomes a coefficient and decreases by one.

The proof relies on two specific facts: \(n\) is a positive integer, so the binomial expansion is a finite sum, and \(x\) is fixed while \(h\) approaches zero. Each remaining term is a fixed coefficient times a positive power of \(h\), and hence tends to zero. No assumption that \(x\) is nonzero is needed.

A Common Mistake: Taking the Limit Too Soon

A frequent error is to take the limit of the numerator before dividing by \(h\). For example, at \(x=2\) with \(f(x)=x^3\), the numerator expands as

$$ (2+h)^3-2^3=12h+6h^2+h^3. $$

A mistaken calculation might say that this numerator tends to zero, so the difference quotient must tend to zero. That omits the denominator. The actual quotient is

$$ \frac{12h+6h^2+h^3}{h}=12+6h+h^2, \qquad h\ne0, $$

and therefore its limit is \(12\), not \(0\). Indeed, the formula gives \(3(2)^2=12\). The numerator and denominator both approach zero, so the limit must be evaluated after simplifying the quotient; the numerator alone does not determine it.

Key takeaway: Expand first, cancel the constant term, divide each remaining term by the nonzero increment \(h\), and only then let \(h\) approach zero.

Scope of the Result

The result established here is specifically for \(x^n\) with a positive integer exponent \(n\). The binomial theorem supplies finitely many terms, and the difference quotient can be simplified term by term before taking the limit. The argument does not, by itself, establish a derivative formula for non-integer powers or for sums of powers. Those require their own justifications.

Within its stated scope, the result is complete: for every positive integer \(n\), the function \(x^n\) has a derivative at every real input \(x\), and that derivative is \(nx^{n-1}\). Examples at positive, negative, and zero inputs illustrate that the same argument applies throughout the real line.

Check Your Understanding

Use the difference quotient and the binomial expansion to answer these questions.

  1. For \(f(x)=x^4\), what is the simplified difference quotient at a general input \(x\) before taking the limit?
  2. Why does the \(k=0\) term in the binomial expansion not remain in the difference quotient numerator?
  3. For a term indexed by \(k\ge2\), what power of \(h\) remains after dividing by \(h\), and what happens to it as \(h\to0\)?
  4. Use the definition to find the derivative of \(x^2\) at \(x=-3\), showing the expansion and limit.
  5. Why is it incorrect to conclude that a difference quotient tends to zero merely because its numerator tends to zero?

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