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Proving the Finite Product Rule by Induction

Learn how induction extends the two-factor product rule to any finite number of differentiable factors, and how to apply and check the resulting formula.

Intermediate 9 min read

What You'll Learn

  • State the finite product rule for differentiable functions on a common interval.
  • Read the product notation that omits one factor.
  • Prove the rule by induction, including the inductive step.
  • Apply the formula to products of three and four factors.
  • Use the rule for repeated factors and check results against an expanded product.
  • Identify and correct the mistake of differentiating every factor at once.

From Two Factors to Any Finite Product

A product of two differentiable functions has a derivative given by the product rule: differentiate one factor at a time, leaving the other factor unchanged, and add the resulting terms. A product with many factors follows the same pattern. The challenge is to state that pattern precisely and justify it for every finite number of factors.

Induction provides a natural proof. We verify the formula for one factor, then show that if it works for a product of \(n\) factors, it also works for a product of \(n+1\) factors. The result gives a reliable formula for differentiating products without repeatedly expanding them first.

Notation and the Product Rule

Let \(I\) be an open interval, and let \(f_1,\ldots,f_n\) be functions differentiable on \(I\). Their product is written \(\prod_{j=1}^{n} f_j(x)\). In the formula below, a product with the \(k\)-th factor omitted means that every factor is included except \(f_k(x)\). When no factors remain, the empty product is defined to be \(1\).

Two-factor product rule: If \(u\) and \(v\) are differentiable on an open interval \(I\), then \(uv\) is differentiable on \(I\), and \((uv)'(x)=u'(x)v(x)+u(x)v'(x)\) for every \(x\in I\).

This rule is the starting point for the induction. It says that each of the two factors contributes one term: in that term, the chosen factor is differentiated and the other factor is left unchanged.

Finite Product Rule: If \(f_1,\ldots,f_n\) are differentiable on an open interval \(I\), where \(n\) is a positive integer, then their product is differentiable on \(I\), and
$$ \frac{d}{dx}\left(\prod_{j=1}^{n} f_j(x)\right) = \sum_{k=1}^{n} f_k'(x)\prod_{\substack{1\le j\le n\\j\ne k}} f_j(x). $$

There are \(n\) terms in the sum. In term \(k\), the factor \(f_k\) is differentiated, while all the other factors remain unchanged. For \(n=1\), the product of the factors other than \(f_1\) is the empty product \(1\), so the formula reduces to \(f_1'(x)\).

Proof by Induction

We prove the rule for every positive integer \(n\), for any collection of \(n\) functions differentiable on the same open interval \(I\). The proof relies on the two-factor product rule stated above.

1
Base case: one factor.
For \(n=1\), the product is \(f_1(x)\). The right-hand side of the claimed formula is \(f_1'(x)\) multiplied by an empty product, which equals \(1\). Thus the formula gives \(f_1'(x)\), as required.
2
Inductive hypothesis.
Fix a positive integer \(n\). Assume that for any \(n\) functions differentiable on \(I\), the derivative of their product is the sum of the \(n\) terms in the finite product formula.
3
Inductive step: add one factor.
Take \(n+1\) differentiable functions \(f_1,\ldots,f_{n+1}\) on \(I\). Write the product of the first \(n\) factors as \(G(x)=\prod_{j=1}^{n}f_j(x)\). By the inductive hypothesis, \(G\) is differentiable and its derivative is given by the formula for \(n\) factors.

The full product is \(G(x)f_{n+1}(x)\). Both functions in this two-factor product are differentiable on \(I\), so the two-factor product rule applies. Substitute the inductive-hypothesis expression for \(G'(x)\):

$$ \begin{aligned} \frac{d}{dx}\left(G(x)f_{n+1}(x)\right) &=G'(x)f_{n+1}(x)+G(x)f_{n+1}'(x)\\ &=\left(\sum_{k=1}^{n}f_k'(x) \prod_{\substack{1\le j\le n\\j\ne k}}f_j(x)\right)f_{n+1}(x) +\left(\prod_{j=1}^{n}f_j(x)\right)f_{n+1}'(x). \end{aligned} $$

In every term of the sum, multiplying by \(f_{n+1}(x)\) includes the new factor among the factors that are not differentiated. The final term differentiates \(f_{n+1}(x)\) and leaves all of \(f_1(x),\ldots,f_n(x)\) unchanged. Therefore the expression is exactly the sum of \(n+1\) terms in which one factor at a time is differentiated.

$$ \frac{d}{dx}\left(\prod_{j=1}^{n+1}f_j(x)\right) = \sum_{k=1}^{n+1}f_k'(x) \prod_{\substack{1\le j\le n+1\\j\ne k}}f_j(x). $$

This proves the inductive step. Since the formula holds for \(n=1\), and its truth for \(n\) implies its truth for \(n+1\), mathematical induction proves the finite product rule for every positive integer \(n\).

Worked Applications

Worked Example: Differentiating Three Factors

Differentiate \(y=(x^2+1)(3x-2)(x-1)\). Set \(f_1(x)=x^2+1\), \(f_2(x)=3x-2\), and \(f_3(x)=x-1\). Their derivatives are \(f_1'(x)=2x\), \(f_2'(x)=3\), and \(f_3'(x)=1\).

The finite product rule gives one term for each differentiated factor:

$$ \begin{aligned} y' &=2x(3x-2)(x-1) +3(x^2+1)(x-1) +(x^2+1)(3x-2). \end{aligned} $$

To simplify, first compute \((3x-2)(x-1)=3x^2-5x+2\). The first term is therefore \(2x(3x^2-5x+2)=6x^3-10x^2+4x\). The second term is \(3(x^2+1)(x-1)=3(x^3-x^2+x-1)=3x^3-3x^2+3x-3\). The third term is \((x^2+1)(3x-2)=3x^3-2x^2+3x-2\). Adding these gives

$$ y'=12x^3-15x^2+10x-5. $$

Check by expanding the original product: \((x^2+1)(3x-2)=3x^3-2x^2+3x-2\), so \(y=3x^4-5x^3+5x^2-5x+2\). Differentiating this polynomial gives \(12x^3-15x^2+10x-5\), agreeing with the finite product rule.

Worked Example: A Product of Four Factors

Differentiate \(y=(x+1)(x-1)(x+2)(x-2)\). The derivatives of the four factors, in order, are \(1,1,1,1\). Differentiating one factor at a time gives

$$ \begin{aligned} y'={}&(x-1)(x+2)(x-2) +(x+1)(x+2)(x-2)\\ &+(x+1)(x-1)(x-2) +(x+1)(x-1)(x+2). \end{aligned} $$

Simplify each term. The first is \((x-1)(x^2-4)=x^3-x^2-4x+4\). The second is \((x+1)(x^2-4)=x^3+x^2-4x-4\). The third is \((x^2-1)(x-2)=x^3-2x^2-x+2\). The fourth is \((x^2-1)(x+2)=x^3+2x^2-x-2\). Their sum is

$$ y'=4x^3-10x. $$

For a check, group the original factors in pairs: \(y=(x^2-1)(x^2-4)=x^4-5x^2+4\). Differentiating gives \(y'=4x^3-10x\), as obtained above. At \(x=1\), this derivative equals \(4-10=-6\). In the four-term formula at \(x=1\), the first, third, and fourth terms are zero because they contain \(x-1\); the second is \(2\cdot3\cdot(-1)=-6\). This also checks the value at that point.

Worked Example: Repeated Factors

Consider the product of \(n\) copies of \(x\), where \(n\) is a positive integer. Each factor has derivative \(1\). The finite product rule has \(n\) terms, and in each term one copy of \(x\) is replaced by \(1\), leaving \(n-1\) copies of \(x\). Thus

$$ \frac{d}{dx}\left(\underbrace{x\cdot x\cdots x}_{n\text{ factors}}\right) = \underbrace{x^{n-1}+x^{n-1}+\cdots+x^{n-1}}_{n\text{ terms}} =nx^{n-1}. $$

The original product is \(x^n\). The power rule gives \(\frac{d}{dx}(x^n)=nx^{n-1}\), so it agrees with the result from the finite product rule. For example, when \(n=4\), the four terms are \(x^3+x^3+x^3+x^3=4x^3\), which is the derivative of \(x^4\).

A Common Mistake

A frequent error is to differentiate every factor in one term, or to multiply all the individual derivatives together. Neither operation follows from the finite product rule. For example, for \(y=(x+1)(x+2)\), multiplying the derivatives would give \(1\cdot1=1\), which is not the derivative of the product.

$$ \begin{aligned} \text{Incorrect:}\quad y'&=(x+1)'(x+2)'=1,\\ \text{Correct:}\quad y'&=1\cdot(x+2)+(x+1)\cdot1\\ &=x+2+x+1=2x+3. \end{aligned} $$

Check the correction by expanding \(y=x^2+3x+2\). Its derivative is \(2x+3\). The key is that each term differentiates exactly one factor and retains every other factor unchanged.

Why the Inductive Proof Is Useful

The proof does more than confirm a pattern seen in a few examples. The inductive step explains precisely why adding one factor adds one new term: applying the two-factor product rule to the existing product and the new factor produces the old terms, each multiplied by the new factor, plus one term in which the new factor is differentiated.

The hypotheses matter. Every factor must be differentiable on the common interval where the formula is used. The induction then ensures that their product is differentiable there as well. The rule is useful when factors are easier to differentiate separately than the full product is to expand. It also keeps the structure visible: each summand records which factor was differentiated.

When using the formula, count the factors and the terms. A product of \(n\) factors has \(n\) summands, one for each choice of the differentiated factor. In each summand, all the other factors must still appear. This simple check often catches missing factors or missing terms before any simplification is done.

Check Your Understanding

For each question, focus on how the factors and terms correspond.

  1. State the finite product rule for \(n\) differentiable functions, including the condition on their domains.
  2. In the inductive step, why can the product of the first \(n\) factors be treated as a differentiable function?
  3. How many terms does the formula have for a product of seven factors, and what is differentiated in each term?
  4. For \(y=(x^2+1)(x-3)\), write the product-rule expression for \(y'\) without expanding.
  5. Explain why multiplying the derivatives of all the factors does not give the finite product rule.

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