Why Repeated Factors Need Care
A product such as \((x+1)^2(x-2)\) contains three factor occurrences: \(x+1\), \(x+1\), and \(x-2\). The two copies of \(x+1\) have the same formula, but they occupy separate places in the product. When differentiating, each place contributes a term.
The product rule for two differentiable functions says that \((uv)'=u'v+uv'\). For more factors, the same idea continues: differentiate one factor at a time, leave the other factors unchanged, and add the resulting terms. Repeated factors are a useful setting for seeing why every occurrence matters.
The Product Rule for Several Factors
Suppose \(f_1,f_2,\ldots,f_n\) are differentiable functions on an interval. Their product is differentiable, and its derivative is the sum of \(n\) terms. In the \(i\)th term, \(f_i\) is differentiated and every other factor is left unchanged.
For three factors, this says to form three terms: differentiate the first factor, then the second, then the third. The other two factors remain as they were in each term.
This formula applies when each factor is differentiable on the interval being considered. The notation does not require the factors to be different functions. If two or more factors are identical, each occurrence still gets its own term.
Why the Several-Factor Rule Works
The formula follows by repeatedly applying the two-factor product rule. Here is a proof for three factors, followed by the same reasoning for any finite number of factors.
Write \(f_1f_2f_3\) as \((f_1f_2)f_3\). Since the factors are differentiable, their products are differentiable and the two-factor product rule applies.
\((f_1f_2f_3)'=(f_1f_2)'f_3+(f_1f_2)f_3'\).
Using \((f_1f_2)'=f_1'f_2+f_1f_2'\), obtain \(f_1'f_2f_3+f_1f_2'f_3+f_1f_2f_3'\).
For \(n\) factors, use induction. The formula is the two-factor product rule when \(n=2\). Suppose it holds for \(n\) factors and let \(P=f_1f_2\cdots f_n\). Then the product of \(n+1\) factors is \(P f_{n+1}\), so
By the induction assumption, \(P'\) is the sum of the \(n\) terms obtained by differentiating each of \(f_1,\ldots,f_n\) in turn. Multiplying that sum by \(f_{n+1}\) appends the unchanged last factor to each of those terms. The remaining term \(P f_{n+1}'\) differentiates the last factor and leaves the first \(n\) factors unchanged. These are exactly the \(n+1\) terms in the stated formula, proving it for every finite \(n\).
A Reliable Procedure
When differentiating a product with repeated factors, first write down every occurrence, even if several entries look alike. Then make one term for each occurrence: differentiate that factor and copy all the other factors unchanged. Only after all terms have been written should you combine identical terms or factor common expressions.
For example, the product \((x+1)^2(x-2)\) should initially be read as \((x+1)(x+1)(x-2)\), not as a list containing only two distinct formulas. There are three factor occurrences, so there will be three product-rule terms.
Worked Examples
Worked Example: Differentiate a Squared Factor Times a Linear Factor
Differentiate \(f(x)=(x+1)^2(x-2)\) by treating the repeated factor as two occurrences.
First write the product as \(f(x)=(x+1)(x+1)(x-2)\). The derivatives of the three factors are \(1\), \(1\), and \(1\). Applying the three-factor rule gives one term for each occurrence:
Check by expanding the original function:
Differentiating this polynomial gives \(3x^2-3\). The product-rule answer also gives \(3(x+1)(x-1)=3(x^2-1)=3x^2-3\), so the results agree.
Worked Example: A Repeated Linear Factor with a Nonunit Derivative
Differentiate \(g(x)=(2x-1)^2(x+3)\). Write the product as \((2x-1)(2x-1)(x+3)\). The derivatives of these factor occurrences are \(2\), \(2\), and \(1\), respectively.
For a check, expand first:
Its derivative is \(12x^2+16x-11\). Expanding the product-rule answer gives \((2x-1)(6x+11)=12x^2+16x-11\), confirming the result. Notice that each occurrence of \(2x-1\) has derivative \(2\); the two matching contributions therefore add to \(4(2x-1)(x+3)\).
Worked Example: Two Different Repeated Factors
Differentiate \(h(x)=(x-1)^3(x+2)^2\). There are five factor occurrences: three copies of \(x-1\), followed by two copies of \(x+2\). Each linear factor has derivative \(1\).
Applying the product rule to all five occurrences gives three equal terms from differentiating a copy of \(x-1\), and two equal terms from differentiating a copy of \(x+2\):
Check by expanding the original function. First, \((x-1)^3=x^3-3x^2+3x-1\) and \((x+2)^2=x^2+4x+4\). Their product is
Therefore \(h'(x)=5x^4+4x^3-15x^2-2x+8\). Expanding the factored answer gives \((x^3-3x+2)(5x+4)=5x^4+4x^3-15x^2-2x+8\), as required.
What Repetition Does to the Terms
If the same differentiable factor \(u(x)\) appears \(m\) times, with another differentiable factor \(v(x)\), the product can be read as \(m\) separate copies of \(u\) followed by \(v\). Differentiating each copy in turn produces \(m\) identical contributions. Differentiating \(v\) produces one further term. Thus, for a positive integer \(m\), the product rule gives
This formula is a consequence of counting the product-rule terms: there are \(m\) choices of a copy of \(u\) to differentiate, and each choice gives the same expression. It is valid wherever \(u\) and \(v\) are differentiable. It is not a reason to omit the individual contributions when first applying the product rule; writing those contributions out can help prevent errors.
A Common Mistake: Counting a Repeated Factor Only Once
For \(f(x)=(x+1)^2(x-2)\), an incorrect calculation might write only
This differentiates one copy of \(x+1\), then differentiates \(x-2\), but it never differentiates the second copy of \(x+1\). The missing term is another \((x+1)(x-2)\). The corrected calculation is
A practical safeguard is to count the terms before simplifying. A product written as \(n\) factor occurrences must initially produce \(n\) product-rule terms. Some terms may be equal, but none should be discarded before they have been accounted for.
Why This Method Is Useful
Writing a product as its visible factors makes the product rule systematic, especially when a factor repeats several times. It shows exactly where each contribution comes from and makes the multiplicity of matching terms clear. The terms can then be combined or factored to make the final derivative easier to read.
When the original expression is a polynomial, expanding and differentiating provides a useful independent check, as in the examples. Expansion is not necessary for the product rule itself; it is simply one way to verify the simplified result. The essential habit is to preserve every factor occurrence until its derivative contribution has been included.
Check Your Understanding
For each question, keep track of factor occurrences before simplifying.
- How many product-rule terms arise from \((x-4)^3(x+1)\) before any terms are combined?
- Differentiate \((x+2)^2(x-1)\) by writing out its three factor occurrences.
- In differentiating \((3x+1)^2(x-2)\), what derivative belongs to each occurrence of \(3x+1\)?
- Why do three identical copies of \(u(x)\) in a product give three matching terms when one copy is differentiated at a time?
- Expand \((x+1)^2(x-2)\) and use the polynomial derivative to check the product-rule result.
