Why Product-Rule Errors Are Easy to Miss
A product has two factors, and differentiating it requires two contributions. A common error is to remember that both factors somehow matter, but lose track of which factor is differentiated in each contribution. The resulting expression may look plausible while being structurally wrong.
This tutorial focuses on three related mistakes: omitting an undifferentiated factor, differentiating both factors in the same term, and writing only one of the two required terms. The key is not to memorize a symbol pattern in isolation. It is to track the two ways a small change in a product can occur: the first factor can change while the second is held in place, or the second can change while the first is held in place.
The Product Rule and Its Terms
In the first term, \(u\) is differentiated and \(v\) remains unchanged: \(u'v\). In the second term, \(v\) is differentiated and \(u\) remains unchanged: \(uv'\). The two terms are added. Neither term is a product of the two derivatives \(u'v'\).
The rule applies when both factors are differentiable at the point under consideration. If the factors are functions of \(x\), each prime means the derivative with respect to \(x\). If a factor is itself a composite function, its derivative must also be computed correctly; the product rule does not replace the chain rule.
Why There Are Two Terms
The reason for the two terms can be seen directly from the difference quotient. Let \(f\) and \(g\) be differentiable at \(x\). Add and subtract \(f(x+h)g(x)\) in the numerator of the difference quotient for their product. This separates the change in the product into a change in \(g\) and a change in \(f\).
As \(h\) tends to \(0\), differentiability of \(g\) gives \(\frac{g(x+h)-g(x)}{h}\to g'(x)\), and differentiability of \(f\) gives \(\frac{f(x+h)-f(x)}{h}\to f'(x)\). Differentiability also implies continuity, so \(f(x+h)\to f(x)\). The other factor \(g(x)\) is already fixed in the second term. Therefore the limit is
This proves the product rule and explains the accounting behind it: the product can change because either factor changes. Each contribution keeps the other factor present.
A Reliable Term-by-Term Check
Before simplifying a product-rule derivative, label the factors \(u\) and \(v\), then write the two terms separately. Check each term against its source: in \(u'v\), only \(u\) is differentiated; in \(uv'\), only \(v\) is differentiated. Finally, check that the terms are added and that no factor has disappeared.
Write the original expression as \(u(x)v(x)\), preserving the entire factors.
Form \(u'(x)v(x)\) and \(u(x)v'(x)\); keep the other factor unchanged in each term.
Combine the two terms only after both have been written, then check the result independently when possible.
Worked Examples
Worked Example: Repairing an Omitted Factor
Differentiate \(y=(x^2+1)(3x-2)\). Let \(u=x^2+1\) and \(v=3x-2\). Then \(u'=2x\) and \(v'=3\). The product rule gives
A tempting incorrect line is \((2x)(3x-2)+3\). It omits \(x^2+1\) from the second term: when \(v\) is differentiated, \(u\) must remain as a factor. To check the answer, expand the original function first:
This agrees with the product-rule result.
Worked Example: Differentiating Both Factors in One Term
Differentiate \(y=(x^2+1)(x^3-2x)\). Set \(u=x^2+1\) and \(v=x^3-2x\). Their derivatives are \(u'=2x\) and \(v'=3x^2-2\). Apply the rule, keeping the undifferentiated factor in each term:
A wrong calculation might replace the two required terms by \(u'v'\), giving \((2x)(3x^2-2)=6x^3-4x\). That expression differentiates both factors in the same term and leaves neither original factor in place. It is not the product rule. For a check, expand the original product:
The expanded-form derivative confirms the result.
Worked Example: Restoring a Missing Product-Rule Term
Differentiate \(y=(x^2-1)(2x+5)\). Choose \(u=x^2-1\), \(v=2x+5\), so \(u'=2x\) and \(v'=2\). Both contributions are required:
If only the first term \(u'v=(2x)(2x+5)\) is written, the second contribution \(uv'=(x^2-1)(2)\) is missing. The product rule does not ask us to choose which factor changes; it accounts for both. Check by expanding the original expression:
This matches the derivative from the product rule.
Spotting the Three Mistakes
These errors can be distinguished by asking what is absent or misplaced in the written expression. The rule always has the same structure, even when the factors are complicated.
| Mistake | What to look for | Correction |
|---|---|---|
| Omitted factor | A term contains a derivative but has lost the other original factor. | Restore the factor that is meant to remain undifferentiated. |
| Both factors differentiated in one term | A term looks like \(u'v'\). | Replace it with the two contributions \(u'v\) and \(uv'\). |
| One required term missing | Only \(u'v\) or only \(uv'\) appears. | Write the other contribution and add it before simplifying. |
Checking only the final simplified algebra may not reveal how an error arose. A structural check is often faster: identify the original factors, then verify that the derivative contains one term for each factor being differentiated, with the other factor preserved. For a quick second check, expand the product first when expansion is manageable and differentiate the resulting sum.
Why the Factor Tracking Matters
The product rule is not the instruction “differentiate each factor and multiply the answers.” It describes how the product changes as \(x\) changes. If the first factor changes slightly, its change is scaled by the current value of the second factor. If the second factor changes, its change is scaled by the current value of the first. These are separate contributions, and both are needed.
For that reason, do not simplify too early. Writing \(u'v+uv'\) explicitly makes it possible to inspect the work before algebra obscures the structure. This habit is especially useful when factors are more involved: the product-rule structure remains the same, while the derivatives of the individual factors may require other differentiation rules.
Check Your Understanding
For each question, focus first on the structure of the product-rule expression.
- If \(y=u(x)v(x)\), which factor remains undifferentiated in the term \(u'(x)v(x)\)?
- A student writes \(u'(x)v'(x)\) as the derivative of \(u(x)v(x)\). Identify the structural mistake and give the correct general expression.
- Differentiate \(y=(x+2)(x^2+1)\), showing both product-rule terms before simplifying.
- For \(y=(x^2+3)(4x-1)\), what term would be missing if a student wrote only \(u'v\)?
- How can expanding the original product provide an independent check of a product-rule derivative?
