Blog · Calculus guides · 18 October 2026 · 4 min read

Product Rule: Formula, Examples & Practice Problems

Q: Why does the product rule work?

Picture a rectangle with sides f and g, so its area is fg. Nudge x a little and the sides grow by Delta f and Delta g. The area gains a strip Delta f· g along one side, a strip f· Delta g along the other, and a tiny corner Delta f Delta g. Divide by Delta x and let it shrink: the strips become f'g and fg', and the corner is so small it vanishes. That is the whole rule.

Q: Does the order matter in the product rule?

No. Because the two terms are added, f'g + fg' and fg' + f'g are the same. That is not true of the quotient rule, where there is a minus sign and the order matters a lot.

Q: When do you use the product rule instead of the chain rule?

Product rule when two functions are multiplied (x cos x); chain rule when one is inside the other (cos(x^2)). Ask yourself whether you would multiply or substitute to evaluate it at a number.

The product rule differentiates two functions multiplied together: $\frac{d}{d x} [f (x) g (x)] = f^{'} (x) g (x) + f (x) g^{'} (x)$ . Differentiate the first and keep the second, then add the first times the derivative of the second. For example, $\frac{d}{d x} [x^{2} sin x] = 2 x sin x + x^{2} cos x$ .

The formula

\frac{d}{d x} [f g] = f^{'} g + f g^{'} or (uv)^{'} = u^{'} v + u v^{'}

Each factor takes a turn being differentiated while the other one waits. Very fair, very civilized, and nothing like what you would guess, which is $f^{'} g^{'}$ .

When to use it

Use the product rule when the function is two (or more) separate pieces multiplied together, and both of them contain $x$ : $x^{2} sin x$ , $x e^{x}$ , $(2 x + 1) (x^{3} - 4)$ . If one factor is just a number, you do not need it, because $\frac{d}{d x} [5 sin x]$ is simply $5 cos x$ . And if one function is sitting inside another, as in $sin (x^{2})$ , that is a composition, which is the chain rule, not a product. Plenty of real problems need both at once.

How to do it, step by step

Name the two factors. Call the first one $f$ and the second one $g$ . In $x^{2} sin x$ , $f = x^{2}$ and $g = sin x$ .
Differentiate each one separately. Write down $f^{'}$ and $g^{'}$ on their own line before you assemble anything: $f^{'} = 2 x$ , $g^{'} = cos x$ .
Assemble $f^{'} g + f g^{'}$ . Two terms, each with exactly one dashed factor.
Simplify. Expand, collect like terms, or factor out anything common. With polynomials, expanding usually gives the tidiest answer; with exponentials and trig, factoring out the shared piece usually does.

Worked examples

Example 1: Two polynomial brackets

Easy

Differentiate:

y = (2 x + 1) (x^{3} - 4)

Here $f = 2 x + 1$ and $g = x^{3} - 4$ , so $f^{'} = 2$ and $g^{'} = 3 x^{2}$ .

\frac{d y}{d x} = 2 (x^{3} - 4) + (2 x + 1) \cdot 3 x^{2}

Expand both terms and collect:

= 2 x^{3} - 8 + 6 x^{3} + 3 x^{2}

You could also have multiplied out the brackets first: $(2 x + 1) (x^{3} - 4) = 2 x^{4} + x^{3} - 8 x - 4$ , and the power rule gives $8 x^{3} + 3 x^{2} - 8$ straight away. Same answer, which makes it a good way to check yourself. It only works for polynomials, though; you cannot expand your way out of $x^{2} sin x$ .

Answer

\frac{d y}{d x} = 8 x^{3} + 3 x^{2} - 8

Example 2: A power times a trig function

Medium

Differentiate:

y = x^{2} sin x

$f = x^{2}$ and $g = sin x$ , so $f^{'} = 2 x$ and $g^{'} = cos x$ . There is no expanding to be done this time; the two terms just sit side by side.

\frac{d y}{d x} = 2 x \cdot sin x + x^{2} \cdot cos x

If you want it factored, $x (2 sin x + x cos x)$ is the same thing.

Answer

\frac{d y}{d x} = 2 x sin x + x^{2} cos x

Example 3: Three factors at once

Harder

Differentiate:

y = x e^{x} cos x

With three factors the rule extends the obvious way: each factor gets one turn being differentiated while the other two stay put.

(f g h)^{'} = f^{'} g h + f g^{'} h + f g h^{'}

Here $f = x$ , $g = e^{x}$ and $h = cos x$ , so $f^{'} = 1$ , $g^{'} = e^{x}$ and $h^{'} = - sin x$ .

\frac{d y}{d x} = 1 \cdot e^{x} cos x + x \cdot e^{x} \cdot cos x + x \cdot e^{x} \cdot (- sin x)

Three factors, three terms: the count matches, so nothing has gone missing. Every term has an $e^{x}$ in it, so pull it out to the front.

Answer

\frac{d y}{d x} = e^{x} (cos x + x cos x - x sin x)

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Differentiate:

y = (3 x - 2) (x^{2} + 5)

Problem 2

Medium

Differentiate:

y = x^{3} e^{x}

Problem 3

Harder

Differentiate:

y = (x^{2} + 1) cos (2 x)

Common mistakes

Multiplying the derivatives. $\frac{d}{d x} [x^{2} sin x]$ is not $2 x cos x$ . The derivative of a product is not the product of the derivatives, however much it looks like it should be.
Using the product rule on a constant times a function. $3 x^{4}$ is not a product that needs the rule. Constants just come along for the ride: $12 x^{3}$ .
Losing a term. The answer always has as many terms as there are factors, before you simplify. Two factors, two terms; three factors, three.
Missing a chain rule inside. In $x^{2} e^{3 x}$ , the derivative of $e^{3 x}$ is $3 e^{3 x}$ , so the answer is $2 x e^{3 x} + 3 x^{2} e^{3 x}$ . The product rule tells you where things go, but each piece still needs its own correct derivative.

Practice it now

Five minutes, no account

Product rule problems show up early in a Sprint, around difficulty 3 and 4, as a linear bracket times a quadratic or cubic one. Expand first or use the rule; the game checks your answer as maths, so any correct form counts.

Play a Sprint See it on the formula sheet

FAQ

What is the product rule in simple terms?: Derivative of the first times the second, plus the first times the derivative of the second. Each factor gets differentiated once, never both at the same time.
Why does the product rule work?: Picture a rectangle with sides $f$ and $g$ , so its area is $f g$ . Nudge $x$ a little and the sides grow by $Δ f$ and $Δ g$ . The area gains a strip $Δ f \cdot g$ along one side, a strip $f \cdot Δ g$ along the other, and a tiny corner $Δ f Δ g$ . Divide by $Δ x$ and let it shrink: the strips become $f^{'} g$ and $f g^{'}$ , and the corner is so small it vanishes. That is the whole rule.
Does the order matter in the product rule?: No. Because the two terms are added, $f^{'} g + f g^{'}$ and $f g^{'} + f^{'} g$ are the same. That is not true of the quotient rule, where there is a minus sign and the order matters a lot.
When do you use the product rule instead of the chain rule?: Product rule when two functions are multiplied ( $x cos x$ ); chain rule when one is inside the other ( $cos (x^{2})$ ). Ask yourself whether you would multiply or substitute to evaluate it at a number.
Is the product rule easier than the quotient rule?: For most people, yes. In our answer data, players get product-rule problems right 84% of the time and quotient-rule problems 76% of the time. That gap is mostly signs.
Where can I find product rule practice problems?: Right on this page: "Your turn" has three product rule practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

When to use it

Use the product rule when the function is two (or more) separate pieces multiplied together, and both of them contain

x

x^{2} sin x

x e^{x}

(2 x + 1) (x^{3} - 4)

. If one factor is just a number, you do not need it, because

\frac{d}{d x} [5 sin x]

is simply

5 cos x

. And if one function is sitting inside another, as in

sin (x^{2})

, that is a composition, which is the chain rule, not a product. Plenty of real problems need both at once.

How to do it, step by step

Name the two factors. Call the first one

f

and the second one

g

. In

x^{2} sin x

f = x^{2}

and

g = sin x

Differentiate each one separately. Write down

f^{'}

and

g^{'}

on their own line before you assemble anything:

f^{'} = 2 x

g^{'} = cos x

Assemble $f^{'} g + f g^{'}$ . Two terms, each with exactly one dashed factor.

Simplify. Expand, collect like terms, or factor out anything common. With polynomials, expanding usually gives the tidiest answer; with exponentials and trig, factoring out the shared piece usually does.

Worked examples

Example 1: Two polynomial brackets

Easy

Differentiate:

y = (2 x + 1) (x^{3} - 4)

Here $f = 2 x + 1$ and $g = x^{3} - 4$ , so $f^{'} = 2$ and $g^{'} = 3 x^{2}$ .

\frac{d y}{d x} = 2 (x^{3} - 4) + (2 x + 1) \cdot 3 x^{2}

Expand both terms and collect:

= 2 x^{3} - 8 + 6 x^{3} + 3 x^{2}

Answer

\frac{d y}{d x} = 8 x^{3} + 3 x^{2} - 8

Example 2: A power times a trig function

Medium

Differentiate:

y = x^{2} sin x

$f = x^{2}$ and $g = sin x$ , so $f^{'} = 2 x$ and $g^{'} = cos x$ . There is no expanding to be done this time; the two terms just sit side by side.

\frac{d y}{d x} = 2 x \cdot sin x + x^{2} \cdot cos x

If you want it factored, $x (2 sin x + x cos x)$ is the same thing.

Answer

\frac{d y}{d x} = 2 x sin x + x^{2} cos x

Example 3: Three factors at once

Harder

Differentiate:

y = x e^{x} cos x

With three factors the rule extends the obvious way: each factor gets one turn being differentiated while the other two stay put.

(f g h)^{'} = f^{'} g h + f g^{'} h + f g h^{'}

Here $f = x$ , $g = e^{x}$ and $h = cos x$ , so $f^{'} = 1$ , $g^{'} = e^{x}$ and $h^{'} = - sin x$ .

\frac{d y}{d x} = 1 \cdot e^{x} cos x + x \cdot e^{x} \cdot cos x + x \cdot e^{x} \cdot (- sin x)

Three factors, three terms: the count matches, so nothing has gone missing. Every term has an $e^{x}$ in it, so pull it out to the front.

Answer

\frac{d y}{d x} = e^{x} (cos x + x cos x - x sin x)

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Differentiate:

y = (3 x - 2) (x^{2} + 5)

Problem 2

Medium

Differentiate:

y = x^{3} e^{x}

Problem 3

Harder

Differentiate:

y = (x^{2} + 1) cos (2 x)

Common mistakes

Multiplying the derivatives.

\frac{d}{d x} [x^{2} sin x]

is not

2 x cos x

. The derivative of a product is not the product of the derivatives, however much it looks like it should be.

Using the product rule on a constant times a function.

3 x^{4}

is not a product that needs the rule. Constants just come along for the ride:

12 x^{3}

Losing a term. The answer always has as many terms as there are factors, before you simplify. Two factors, two terms; three factors, three.

Missing a chain rule inside. In

x^{2} e^{3 x}

, the derivative of

e^{3 x}

3 e^{3 x}

, so the answer is

2 x e^{3 x} + 3 x^{2} e^{3 x}

. The product rule tells you where things go, but each piece still needs its own correct derivative.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

What is the product rule in simple terms?

Derivative of the first times the second, plus the first times the derivative of the second. Each factor gets differentiated once, never both at the same time.

Why does the product rule work?

Picture a rectangle with sides

f

and

g

, so its area is

f g

. Nudge

x

a little and the sides grow by

Δ f

and

Δ g

. The area gains a strip

Δ f \cdot g

along one side, a strip

f \cdot Δ g

along the other, and a tiny corner

Δ f Δ g

. Divide by

Δ x

and let it shrink: the strips become

f^{'} g

and

f g^{'}

, and the corner is so small it vanishes. That is the whole rule.

Does the order matter in the product rule?

No. Because the two terms are added,

f^{'} g + f g^{'}

and

f g^{'} + f^{'} g

are the same. That is not true of the quotient rule, where there is a minus sign and the order matters a lot.

When do you use the product rule instead of the chain rule?

Product rule when two functions are multiplied (

x cos x

); chain rule when one is inside the other (

cos (x^{2})

). Ask yourself whether you would multiply or substitute to evaluate it at a number.

Is the product rule easier than the quotient rule?

For most people, yes. In our answer data, players get product-rule problems right 84% of the time and quotient-rule problems 76% of the time. That gap is mostly signs.

Where can I find product rule practice problems?

Right on this page: "Your turn" has three product rule practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Product Rule: Formula, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Two polynomial brackets

Example 2: A power times a trig function

Example 3: Three factors at once

Your turn

Common mistakes

FAQ

Keep going

Product Rule: Formula, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Two polynomial brackets

Example 2: A power times a trig function

Example 3: Three factors at once

Your turn

Common mistakes

FAQ

Keep going