Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
When two factors are multiplied, the derivative is a sum of two terms: dxd[f(x)g(x)]=f′(x)g(x)+f(x)g′(x). The structure has a reason. A product can change because the first factor moved while the second held still, or because the second moved while the first held still, and those are two separate stories that have to be added. The rule is symmetric, so the order of the two terms never matters, but neither term can be dropped. The errors are predictable: multiplying the two derivatives instead of adding the two cross products, or writing only one of the two terms. Three factors work the same way with three terms, one per factor, which is the fastest way to see why the sum is the shape it is.
Figure 1 — Growing both factors at once grows the rectangle in three pieces. The two strips are the two terms of the product rule, and the small corner carries a product of two increments, which shrinks faster than either strip and disappears in the limit.
dxd[fg]=f′g+fg′
The two terms are mirror images, so writing them in either order is equally correct. That symmetry is a convenience, not a licence to write one term and stop. A useful check: if f is constant, the first term vanishes and the rule collapses to the constant multiple rule. If both factors are x, the rule gives 1⋅x+x⋅1=2x, which matches the power rule. The product rule has to agree with the simpler rules in those cases, and it does.
For three factors, one term appears for each factor that is allowed to change:
dxd[fgh]=f′gh+fg′h+fgh′
The pattern generalises in the obvious way, and it is worth seeing once because it makes the two-factor case feel less arbitrary. Each term differentiates exactly one factor and holds the rest fixed.
| Expression | Product rule needed? | Why |
|---|---|---|
| x2⋅x3 | No | Combine to x5 first |
| xex | Yes | Unlike bases, no combining |
| xsinx | Yes | A polynomial times a trigonometric function |
| (x+1)(x−1) | No | Expand to x2−1 first |
| lnx⋅cosx | Yes | Nothing simplifies |
Half of the product-rule questions on the exam can be avoided by simplifying first, and choosing the shorter route is part of the skill. The rule is for products that cannot be combined, not for every multiplication sign.
Borrowed idea: Which factor changes first? Take them one at a time. Hold the first factor still, differentiate the second, then swap the roles and add the two results. Does the order matter? No, the sum comes out the same either way, so pick whichever order makes the algebra shorter. What goes wrong most often is the shortcut that multiplies the two derivatives instead of adding the cross terms. That shortcut fails because a product changes in two independent ways, and each way deserves its own term. Have you written both terms before simplifying any of them? Then check that your answer has exactly two pieces, since an answer with one piece has lost a term.
| Expression | Correct derivative | The error to expect |
|---|---|---|
| xex | ex+xex | ex alone |
| xsinx | sinx+xcosx | cosx alone |
| excosx | excosx−exsinx | ex(−sinx) alone |
| (2x+1)(x2) | 2x2+(2x+1)(2x) | 2⋅2x |
| x2ex | 2xex+x2ex | 2xex alone |
① Differentiate xex with the product rule, then check the answer at x=0 against a numerical derivative. Two independent methods agreeing is worth more than a memorised formula.
② Take the same function, differentiate it by expanding first where possible, and confirm the two routes agree. On (x+1)(x−1) the expansion route is faster, which is the point.
③ Write out the three-factor rule once from memory and check it against the pattern. If you can produce three terms, the two-factor case stops feeling arbitrary.
Core Archetype (Simplest Form): Differentiate f(x)=x2ex.
Standard Execution:
- Name the factors. f1=x2 and f2=ex.
- Differentiate each. f1′=2x and f2′=ex.
- Cross and add. f′(x)=2xex+x2ex.
- Factor if asked. f′(x)=xex(2+x).
Exam Variation (Context / Parameter Shift): A population is modeled by P(t)=t2e−0.1t for t≥0, in thousands, with t in years. Find P′(t) and determine whether the population is increasing at t=5.
Solution: With f=t2 and g=e−0.1t, the product rule gives P′(t)=2te−0.1t+t2(−0.1)e−0.1t=e−0.1t(2t−0.1t2). At t=5 this is e−0.5(10−2.5)=7.5e−0.5>0, so the population is increasing.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
If f(x)=xex, what is f′(x)?
MCQ 2
If g(x)=xcosx, what is g′(x)?
MCQ 3
Which expression is best differentiated by simplifying first rather than by the product rule?
A population of fish in a lake is modeled by P(t)=t2e−0.1t for 0≤t≤30, measured in thousands, where t is the number of years since the model began.
(a) Find P′(t).
(b) Find P′(5) and state, with units, what that number means for the population.
(c) Find every time in 0<t<30 at which P′(t)=0, and interpret the largest of them.
(d) Explain why the product rule is unavoidable here, unlike in a model such as Q(t)=t2⋅t3.
| Move | |
|---|---|
| fg | f′g+fg′ |
| Three factors | f′gh+fg′h+fgh′ |
| Same base | Combine the powers first |
| Expandable product | Expand first, then differentiate |
| One factor constant | Falls back to the constant multiple rule |
| Check | Both terms present, and each differentiates exactly one factor |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
When two factors are multiplied, the derivative is a sum of two terms: dxd[f(x)g(x)]=f′(x)g(x)+f(x)g′(x). The structure has a reason. A product can change because the first factor moved while the second held still, or because the second moved while the first held still, and those are two separate stories that have to be added. The rule is symmetric, so the order of the two terms never matters, but neither term can be dropped. The errors are predictable: multiplying the two derivatives instead of adding the two cross products, or writing only one of the two terms. Three factors work the same way with three terms, one per factor, which is the fastest way to see why the sum is the shape it is.
Figure 1 — Growing both factors at once grows the rectangle in three pieces. The two strips are the two terms of the product rule, and the small corner carries a product of two increments, which shrinks faster than either strip and disappears in the limit.
dxd[fg]=f′g+fg′
The two terms are mirror images, so writing them in either order is equally correct. That symmetry is a convenience, not a licence to write one term and stop. A useful check: if f is constant, the first term vanishes and the rule collapses to the constant multiple rule. If both factors are x, the rule gives 1⋅x+x⋅1=2x, which matches the power rule. The product rule has to agree with the simpler rules in those cases, and it does.
For three factors, one term appears for each factor that is allowed to change:
dxd[fgh]=f′gh+fg′h+fgh′
The pattern generalises in the obvious way, and it is worth seeing once because it makes the two-factor case feel less arbitrary. Each term differentiates exactly one factor and holds the rest fixed.
| Expression | Product rule needed? | Why |
|---|---|---|
| x2⋅x3 | No | Combine to x5 first |
| xex | Yes | Unlike bases, no combining |
| xsinx | Yes | A polynomial times a trigonometric function |
| (x+1)(x−1) | No | Expand to x2−1 first |
| lnx⋅cosx | Yes | Nothing simplifies |
Half of the product-rule questions on the exam can be avoided by simplifying first, and choosing the shorter route is part of the skill. The rule is for products that cannot be combined, not for every multiplication sign.
Borrowed idea: Which factor changes first? Take them one at a time. Hold the first factor still, differentiate the second, then swap the roles and add the two results. Does the order matter? No, the sum comes out the same either way, so pick whichever order makes the algebra shorter. What goes wrong most often is the shortcut that multiplies the two derivatives instead of adding the cross terms. That shortcut fails because a product changes in two independent ways, and each way deserves its own term. Have you written both terms before simplifying any of them? Then check that your answer has exactly two pieces, since an answer with one piece has lost a term.
| Expression | Correct derivative | The error to expect |
|---|---|---|
| xex | ex+xex | ex alone |
| xsinx | sinx+xcosx | cosx alone |
| excosx | excosx−exsinx | ex(−sinx) alone |
| (2x+1)(x2) | 2x2+(2x+1)(2x) | 2⋅2x |
| x2ex | 2xex+x2ex | 2xex alone |
① Differentiate xex with the product rule, then check the answer at x=0 against a numerical derivative. Two independent methods agreeing is worth more than a memorised formula.
② Take the same function, differentiate it by expanding first where possible, and confirm the two routes agree. On (x+1)(x−1) the expansion route is faster, which is the point.
③ Write out the three-factor rule once from memory and check it against the pattern. If you can produce three terms, the two-factor case stops feeling arbitrary.
Core Archetype (Simplest Form): Differentiate f(x)=x2ex.
Standard Execution:
- Name the factors. f1=x2 and f2=ex.
- Differentiate each. f1′=2x and f2′=ex.
- Cross and add. f′(x)=2xex+x2ex.
- Factor if asked. f′(x)=xex(2+x).
Exam Variation (Context / Parameter Shift): A population is modeled by P(t)=t2e−0.1t for t≥0, in thousands, with t in years. Find P′(t) and determine whether the population is increasing at t=5.
Solution: With f=t2 and g=e−0.1t, the product rule gives P′(t)=2te−0.1t+t2(−0.1)e−0.1t=e−0.1t(2t−0.1t2). At t=5 this is e−0.5(10−2.5)=7.5e−0.5>0, so the population is increasing.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
If f(x)=xex, what is f′(x)?
MCQ 2
If g(x)=xcosx, what is g′(x)?
MCQ 3
Which expression is best differentiated by simplifying first rather than by the product rule?
A population of fish in a lake is modeled by P(t)=t2e−0.1t for 0≤t≤30, measured in thousands, where t is the number of years since the model began.
(a) Find P′(t).
(b) Find P′(5) and state, with units, what that number means for the population.
(c) Find every time in 0<t<30 at which P′(t)=0, and interpret the largest of them.
(d) Explain why the product rule is unavoidable here, unlike in a model such as Q(t)=t2⋅t3.
| Move | |
|---|---|
| fg | f′g+fg′ |
| Three factors | f′gh+fg′h+fgh′ |
| Same base | Combine the powers first |
| Expandable product | Expand first, then differentiate |
| One factor constant | Falls back to the constant multiple rule |
| Check | Both terms present, and each differentiates exactly one factor |
Strengthen this topic with AP-style practice, or continue to next topic.