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Calculus BC · Unit 2 · 2.8
AP Calculus · Differentiation: Definition and Fundamental Properties
The Product Rule
Differentiate one factor at a time, leave the other alone, and add.
key idea$(fg)'=f'g+fg'$
The rule
The derivative of a product is not the product of the derivatives. That single false instinct accounts for most errors on this topic, so the rule is worth stating carefully: differentiate each factor in turn, leaving the other one alone, then add.
The Product Rule
\[\frac{d}{dx}\bigl[f(x)g(x)\bigr]=f'(x)g(x)+f(x)g'(x)\]
In words: first times the derivative of the second, plus second times the derivative of the first. The order of the two terms does not matter, but each term must contain exactly one derivative and one undifferentiated factor.
Why it is not $f'g'$
Test it on $f(x)=g(x)=x$. The product is $x^{2}$, whose derivative is $2x$. But $f'g'=1\cdot1=1$. The rule gives $1\cdot x + x\cdot 1 = 2x$ — correct. One counterexample settles it.
Pattern recognition
Before differentiating, decide whether you need the product rule at all. Often you do not.
| If the expression looks like… | Do this | Why |
|---|
| $x^{2}\sin x$ — two genuinely different functions multiplied | product rule | neither factor is a constant; both depend on $x$ |
| $5x^{3}$ — a constant times a function | constant multiple rule | the constant is not a function of $x$; $\tfrac{d}{dx}(5x^3)=15x^2$ |
| $x^{2}\cdot x^{5}$ — like bases | simplify first to $x^{7}$ | one power rule beats a product rule |
| $\dfrac{x^{2}}{\sin x}$ — a quotient | quotient rule, or rewrite as $x^{2}(\sin x)^{-1}$ | either works; the quotient rule is usually cleaner |
| $x^{2}\sin x\cos x$ — three factors | product rule twice, or group two factors as one | extend by association, not by guessing |
| $(x^{2}+1)^{4}$ — a power of a function | chain rule, not the product rule | this is a composition, not a product |
The commonest wasted effort is reaching for the product rule on $x^{2}\cdot x^{5}$ or on a constant multiple. Simplify first, differentiate second.
Worked examples
EX 1
The basic pattern
$x^{2}\sin x$Differentiate $y=x^{2}\sin x$.
1
Name the factors
$f=x^{2}$ and $g=\sin x$.
2
Differentiate each
$f'=2x$ and $g'=\cos x$.
3
Assemble
$y'=f'g+fg'=2x\sin x+x^{2}\cos x$.
4
Check the shape
Each term has exactly one derivative and one original factor. If a term contains two primes or none, the assembly is wrong.
EX 2
Simplify before you differentiate
$x^{3}(x^{2}+4)$Differentiate $y=x^{3}(x^{2}+4)$ two ways and compare.
1
Product rule
$f=x^{3}$, $g=x^{2}+4$, so $f'=3x^{2}$, $g'=2x$. Then $y'=3x^{2}(x^{2}+4)+x^{3}(2x)=3x^{4}+12x^{2}+2x^{4}=5x^{4}+12x^{2}$.
2
Expand first instead
$y=x^{5}+4x^{3}$, so $y'=5x^{4}+12x^{2}$.
3
Compare
Identical, as they must be. The second route is faster and has fewer places to slip.
4
Rule of thumb
If expanding is easy, expand. Save the product rule for factors you cannot multiply out, like $\sin x$ or $e^{x}$.
EX 3
Combined with the chain rule
$xe^{3x}$Differentiate $y=xe^{3x}$.
1
Identify the structure
This is a product: $f=x$ and $g=e^{3x}$. The second factor is itself a composition.
2
Differentiate each factor
$f'=1$. For $g$, the chain rule gives $g'=3e^{3x}$ — the inner derivative $3$ comes out front.
3
Apply the product rule
$y'=1\cdot e^{3x}+x\cdot 3e^{3x}=e^{3x}+3xe^{3x}$.
4
Factor the answer
$y'=e^{3x}(1+3x)$. Factoring is worth doing — it is usually the form needed for finding critical points.
5
Common slip
Writing $g'=e^{3x}$ and forgetting the inner $3$. The chain rule applies inside the product rule, not instead of it.
EX 4
Three factors
$x\sin x\cos x$Differentiate $y=x\sin x\cos x$.
1
Group into two factors
Treat it as $f\cdot g$ with $f=x$ and $g=\sin x\cos x$.
2
Differentiate the group
$g$ is itself a product: $g'=\cos x\cos x+\sin x(-\sin x)=\cos^{2}x-\sin^{2}x$.
3
Apply the outer product rule
$y'=1\cdot\sin x\cos x+x\left(\cos^{2}x-\sin^{2}x\right)$.
4
Note the general pattern
For three factors the rule extends to $(fgh)'=f'gh+fg'h+fgh'$ — one term per factor, each with a single derivative.
5
Optional simplification
Using $\sin 2x=2\sin x\cos x$ and $\cos 2x=\cos^{2}x-\sin^{2}x$, this is $\tfrac12\sin 2x+x\cos 2x$.
EX 5
Product rule from a table
tableGiven $f(2)=3$, $f'(2)=-1$, $g(2)=5$, $g'(2)=4$, find $(fg)'(2)$.
1
Write the rule with the numbers in place
$(fg)'(2)=f'(2)g(2)+f(2)g'(2)$.
2
Substitute
$=(-1)(5)+(3)(4)$.
4
Why this form is examined
You never see formulas for $f$ or $g$ — only values. That forces you to use the rule structurally rather than by pattern-matching an expression.
Watch out
$(fg)'\neq f'g'$
The single most common error on this topic. If you are ever unsure, test your version on $f=g=x$: the answer must come out as $2x$.
One prime per term
Each term of the assembled answer contains exactly one differentiated factor. Two primes in a term, or none, means a mis-assembly.
The chain rule still applies inside
In $xe^{3x}$ the factor $e^{3x}$ needs its own inner derivative. Product and chain rules compose; they do not replace one another.
Do not reach for it on constants or like bases
$5x^{3}$ needs the constant multiple rule and $x^{2}\cdot x^{5}$ should be simplified to $x^{7}$ first. Using the product rule is not wrong there, just slower and more error-prone.