Why Products Need More Than One Product-Rule Step
A polynomial may be written as a product of simpler polynomial factors. For two factors, the product rule says that if \(u\) and \(v\) are differentiable, then \((uv)'=u'v+uv'\). With three or more factors, the same idea applies repeatedly: each term accounts for one factor changing while all the other factors remain in place.
This matters because a product of several factors is not differentiated by differentiating every factor simultaneously. Instead, the derivative is a sum. The number of terms in that sum matches the number of factors, and each term has exactly one differentiated factor.
The Rule for Three Factors
Let \(f\), \(g\), and \(h\) be differentiable functions. Apply the product rule first to the product \((fg)h\). The derivative of \(fg\) is \(f'g+fg'\), so the result is:
There are three terms, one for each factor. In the first, \(f\) is differentiated; in the second, \(g\) is differentiated; and in the third, \(h\) is differentiated. The order of the factors is preserved in every term.
Worked Example: Differentiating Three Linear Factors
Worked Example: Differentiating Three Linear Factors
Differentiate \(P(x)=(x+1)(x-2)(x+3)\). Label the factors \(f=x+1\), \(g=x-2\), and \(h=x+3\). Their derivatives are \(f'=1\), \(g'=1\), and \(h'=1\).
Use the three-factor rule, differentiating one factor in each term:
Now simplify each product:
Adding the three results gives:
Check by expanding \(P\) first. Since \((x-2)(x+3)=x^2+x-6\),
so differentiating the expanded polynomial gives \(P'(x)=3x^2+4x-5\), matching the result from the product rule.
Products with Four or More Factors
The same pattern continues when there are more factors. For four differentiable functions, regard the product as \((fgh)k\) and apply the two-factor product rule. The derivative of \(fgh\) already has three terms; multiplying those by \(k\), then adding the term where \(k\) is differentiated, gives four terms.
In general, for differentiable functions \(f_1,f_2,\ldots,f_n\), the derivative is a sum of \(n\) terms. The \(i\)-th term contains \(f_i'\) in place of \(f_i\), with every other factor unchanged.
This compact notation is a precise version of the “one factor at a time” instruction. For example, if \(n=4\), the sum has four terms. The condition \(j\ne i\) means that the factor with index \(i\) is left out of the unchanged product because it has been differentiated.
Why the General Rule Works
The general rule follows by induction from the ordinary two-factor product rule. Induction means proving a starting case and then showing that if the statement holds for a product of \(n\) factors, it also holds for a product of \(n+1\) factors.
For \(n=1\), the derivative of \(f_1\) is \(f_1'\). This is exactly the one-term formula.
Let \(F=f_1f_2\cdots f_n\), and assume \(F'=\sum_{i=1}^{n} f_i'\prod_{j\ne i,\,1\le j\le n}f_j\).
For a further differentiable function \(f_{n+1}\), use the two-factor product rule on \(Ff_{n+1}\): \((Ff_{n+1})'=F'f_{n+1}+Ff_{n+1}'\).
The term \(F'f_{n+1}\) gives the first \(n\) terms, each with the additional unchanged factor \(f_{n+1}\). The term \(Ff_{n+1}'\) gives the final term, where only \(f_{n+1}\) is differentiated.
More explicitly, substituting the induction assumption yields:
The sum on the last line contains the terms for differentiating factors \(1\) through \(n\); the final term differentiates factor \(n+1\). Together they are exactly the stated formula for \(n+1\) factors. Since the rule holds for one factor and each case implies the next, it holds for every positive integer \(n\).
Worked Example: Four Factors
Worked Example: Four Factors
Differentiate \(Q(x)=(x-1)(x+1)(x+2)(x-2)\). The derivative of each linear factor is \(1\), so the four-factor rule gives:
Simplify each term, keeping track of its sign and powers:
Adding like terms, the cubic terms total \(4x^3\), the quadratic terms cancel, the linear terms total \(-10x\), and the constants cancel. Therefore \(Q'(x)=4x^3-10x\).
Check by multiplying the original factors in pairs:
This agrees with the four-term calculation.
Worked Example: A Factor with a Nonconstant Derivative
Worked Example: A Factor with a Nonconstant Derivative
Differentiate \(R(x)=(2x-1)(x^2+1)(x-3)\). Name the factors \(f=2x-1\), \(g=x^2+1\), and \(h=x-3\). Their derivatives are \(f'=2\), \(g'=2x\), and \(h'=1\). Thus:
Simplify each term separately:
Adding the terms gives:
Check by expanding \(R\). First, \((2x-1)(x^2+1)=2x^3-x^2+2x-1\). Then:
Differentiating this polynomial gives \(8x^3-21x^2+10x-7\), confirming the result.
A Common Mistake: Leaving Out a Term
With three factors, a frequent error is to differentiate the first and last factors but forget the middle one. For \(P(x)=(x+1)(x+2)(x+3)\), that incomplete calculation would be:
It omits the term \((x+1)(x+3)\), which comes from differentiating the middle factor. At \(x=0\), the incorrect expression gives \(2\cdot3+1\cdot2=8\). The correct rule gives:
As a check, expanding \(P(x)\) gives \(x^3+6x^2+11x+6\), whose derivative at \(0\) is also \(11\). To avoid this mistake, count the factors before writing the derivative: a product of \(n\) factors must produce \(n\) terms.
Using the Rule Efficiently
For a factored polynomial, keeping the derivative in product-rule form can be useful. It often avoids expanding a large product, and it makes clear which factor contributed each term. If a final answer needs to be simplified, expand or combine terms only after the complete derivative has been written.
- List the factors in their original order.
- Find the derivative of each factor.
- Write one term for each factor, differentiating that factor alone.
- Check that every other factor appears unchanged in that term.
- Simplify and, when practical, verify by expanding the original product.
If one factor is a constant, its derivative is zero, so the term corresponding to that factor contributes zero. The other terms still follow the same pattern. The rule also requires each factor to be differentiable at the input values under consideration; polynomial factors are differentiable for every real \(x\).
Check Your Understanding
For each question, write enough work to show how the factors and terms are accounted for.
- Write the product-rule expression for the derivative of \((x-2)(x+1)(x+4)\), before simplifying.
- How many terms should appear when differentiating a product of six differentiable factors?
- Differentiate \((x+2)(x^2-1)(x-1)\) using the three-factor rule.
- Explain why the derivative of \(fghk\) has four terms rather than one term containing \(f'g'h'k'\).
- Differentiate \((3x+1)(x-2)(x+5)\), and check your result by expanding the original product.
