Why Multi-Factor Products Need a Systematic Rule
A polynomial product may contain several factors, each of which is itself a sum of powers of \(x\). Differentiating the product requires accounting for how every factor changes. With three or more factors, it is easy to omit a term if the calculation is not organized.
The product rule gives a direct method: differentiate one factor at a time, leave the other factors unchanged, and add the resulting terms. Once those terms are written, the sum and power rules let us differentiate the polynomial factors, and ordinary algebra lets us expand and collect the result.
The Rule for Several Factors
For differentiable functions \(f\) and \(g\), the product rule states that
For polynomials, the factors and their derivatives are defined for every real \(x\). The sum rule says that the derivative of a finite sum is the sum of the derivatives, and the power rule says that \(\frac{d}{dx}x^k=kx^{k-1}\) for a nonnegative integer \(k\), with the derivative of a constant equal to zero. These rules provide the factor derivatives needed in each product-rule term.
The notation \(\prod\) means to multiply the indicated factors, and \(\sum\) means to add the indicated terms. For example, with three factors the formula is
Why the Rule Works
The multi-factor formula follows from the two-factor product rule. The argument below proves it for every positive integer number of factors, including the one-factor case.
The product has just one factor, so its derivative is \(f_1'\). In the formula, the product of the other factors is the empty product \(1\), and the sum has one term, \(f_1'\cdot 1=f_1'\). Thus the formula holds for \(n=1\).
Assume the formula holds for a product \(P_n=f_1f_2\cdots f_n\) of \(n\) differentiable factors. Then \(P_n'=\sum_{j=1}^{n}f_j'\prod_{1\le i\le n,\ i\ne j}f_i\).
The product of \(n+1\) factors is \(P_nf_{n+1}\). By the two-factor product rule, its derivative is \(P_n'f_{n+1}+P_nf_{n+1}'\).
Substituting the induction hypothesis gives \(\sum_{j=1}^{n}\left(f_j'\prod_{1\le i\le n,\ i\ne j}f_i\right)f_{n+1}+\left(\prod_{i=1}^{n}f_i\right)f_{n+1}'\). The sum contains the terms where one of the first \(n\) factors is differentiated; the final term differentiates \(f_{n+1}\). Together these are exactly the formula for \(n+1\) factors.
By mathematical induction, the formula holds for every positive integer \(n\). This proof also explains the practical method: use the product rule to add one factor at a time, or write the equivalent list of terms in which each factor is differentiated once.
Worked Example: Differentiating Three Polynomial Factors
Worked Example: Differentiating Three Polynomial Factors
Differentiate \(f(x)=(x^2+1)(2x-3)(x^2-x+2)\), then expand and collect the derivative terms.
Name the factors \(A=x^2+1\), \(B=2x-3\), and \(C=x^2-x+2\). By the power and sum rules, \(A'=2x\), \(B'=2\), and \(C'=2x-1\). The three-factor rule gives
Expand each term separately. First,
For the second term,
For the third term, first multiply \(A\) and \(B\), then multiply by \(2x-1\):
Now add like powers, displaying their coefficients:
Check by expanding the original product. First, \((x^2+1)(2x-3)=2x^3-3x^2+2x-3\). Then
Differentiate this expanded polynomial using the power and sum rules:
This matches the product-rule result.
Worked Example: Four Factors
Worked Example: Four Factors
Differentiate \(g(x)=x(x+1)(x+2)(x+3)\).
There are four factors, so write one term for each differentiated factor:
Expand each term. The first is
The second is
The third is
The fourth is
Add the four results:
Check by expanding the original product. Since \((x+1)(x+2)=x^2+3x+2\),
Differentiating gives \(g'(x)=4x^3+18x^2+22x+6\), agreeing with the four-factor calculation.
Worked Example: Two Polynomial Factors
Worked Example: Two Polynomial Factors
Differentiate \(h(x)=(3x^2-2x+1)(x^3+2x)\).
Let \(A=3x^2-2x+1\) and \(B=x^3+2x\). The power and sum rules give \(A'=6x-2\) and \(B'=3x^2+2\). Apply the two-factor product rule:
Expand the first term:
Expand the second term:
Collect like terms:
To check, expand \(h\) itself:
Its derivative is \(15x^4-8x^3+21x^2-8x+2\), which matches the product-rule result.
A Common Omission: Leaving Out a Factor's Term
For three factors, there must be three terms. In the first worked example, an incomplete calculation might write
and stop there. This differentiates the first factor in one term and the second factor in the other, but it never differentiates the third factor. The omitted term is \((x^2+1)(2x-3)(2x-1)\), which is not generally zero. For example, at \(x=0\) the two written terms sum to \(4\), while the omitted term equals \(3\). The complete derivative therefore has value \(7\) at \(x=0\), consistent with \(10x^4-20x^3+27x^2-22x+7\).
Organizing the Calculation
A reliable sequence keeps the product-rule structure visible until every term is present. Then expand, collect, and check. Expanding the original product first can also provide a check, but it may involve more algebra than applying the product rule directly.
Write the polynomial as a product of its factors and label them if that makes the calculation easier to track.
Use the power and sum rules, keeping constants and signs in place.
Differentiate one factor in each term and copy all the other factors unchanged.
Expand the terms carefully, then combine coefficients of matching powers of \(x\).
Expand the original product and differentiate it, or compare with an independently simplified expression when that is convenient.
This approach matters because a correct derivative is not just a list of terms: the terms must account for every factor's contribution. Keeping the structure visible helps catch omissions before they become buried in expanded algebra.
Check Your Understanding
For each question, write the derivative rule sequence before expanding or simplifying.
- How many terms appear in the product-rule formula for five differentiable factors?
- Write the product-rule expression for the derivative of \((x^2+1)(x-2)(3x+1)\), without expanding it.
- For \(p(x)=x(x+1)(x+2)\), what factor is differentiated in the third term of \(p'(x)\)?
- Why must the derivative formula for \(n=1\) be checked separately in the induction proof?
- What is one way to check a derivative obtained by applying the product rule to polynomial factors?
