The product rule for differentiation

The product rule tells you how to differentiate two functions multiplied together. If y = uv, where u and v are both functions of x, then dy/dx = u dv/dx + v du/dx. In words: the first times the derivative of the second, plus the second times the derivative of the first. You need it for functions such as x² sin x or x e^(3x), where you cannot simply differentiate each part and multiply.

The product rule formula

UK textbooks usually write the product rule with u and v. US textbooks and many calculators use f and g with prime notation. They are the same rule.

Addition does not care about order, so u'v + uv' and uv' + vu' are the same thing. The one thing you must not do is multiply the two derivatives: the derivative of uv is not u'v'. For example, x² × x = x³ has derivative 3x², but 2x × 1 = 2x, which is wrong.

Product rule differentiation in 4 steps

1. Split the function into two factors and call them u and v.
2. Differentiate each one to get du/dx and dv/dx. Use the chain rule here if either factor is itself a function of a function.
3. Substitute into dy/dx = u dv/dx + v du/dx.
4. Simplify, usually by taking out common factors. Exam questions often ask for the answer in a factorised form.

Laying out the four pieces in a small grid keeps the working tidy and is what most mark schemes expect to see:

When to use the product rule

Use it when the function is a product of two parts that both contain x. If one factor is just a number, such as 5 sin x, you do not need it: the number stays in front. If the product is easy to multiply out, such as x²(x + 1) = x³ + x², expanding first is quicker.

To learn all three rules together, the Calculus course covers the chain, product and quotient rules, with a practice that mixes them.

Product rule examples

Example 1: a polynomial times a trig function

Differentiate y = x³ sin x.

Let u = x³ and v = sin x. Then du/dx = 3x² and dv/dx = cos x.

Both terms contain x², so take it out as a common factor.

Example 2: an exponential with the chain rule inside

Differentiate y = x² e^(3x) and write the answer in its simplest factorised form.

Let u = x² and v = e^(3x). Then du/dx = 2x, and the chain rule gives dv/dx = 3e^(3x).

Example 3: finding a stationary point

Find the stationary point of y = x ln x, for x > 0.

Let u = x and v = ln x. Then du/dx = 1 and dv/dx = 1/x.

The first term simplifies to 1, so the gradient function is 1 + ln x. The gradient is zero at a stationary point. Setting 1 + ln x equal to zero gives ln x = −1, which means x = e⁻¹ = 1/e. Then y = (1/e) × ln(1/e) = −1/e. The stationary point is (1/e, −1/e), about (0.368, −0.368).

Here x is the input and y = x ln x, where log means the natural logarithm. The curve dips to its lowest point at x = 1/e, exactly where the product rule says the gradient is zero, then rises through (1, 0).

Example 4: the equation of a tangent

Find the equation of the tangent to y = e^(2x) cos x at the point where x = 0.

Let u = e^(2x) and v = cos x. Then du/dx = 2e^(2x) and dv/dx = −sin x.

At x = 0, y = e⁰ cos 0 = 1 and the gradient is e⁰(2 − 0) = 2. The tangent passes through (0, 1) with gradient 2, so its equation is y = 2x + 1.

Example 5: a factorised answer, A-level style

Given y = x³(2x + 5)⁴, show that dy/dx = x²(2x + 5)³(14x + 15).

Let u = x³ and v = (2x + 5)⁴. Then du/dx = 3x², and the chain rule gives dv/dx = 4(2x + 5)³ × 2 = 8(2x + 5)³.

Both terms contain x² and (2x + 5)³. Take those out:

Take out the lowest power of each common factor. Here that is x² and (2x + 5)³. Expanding everything would be far longer and easy to get wrong.

Example 6: three functions multiplied

Differentiate y = x e^x sin x.

For three factors, differentiate one at a time and keep the other two, then add the three terms: (uvw)′ = u′vw + uv′w + uvw′.

You get the same answer by treating x e^x as u and sin x as v, then using the product rule again to find du/dx.

Try the product rule yourself

This practice from the Calculus course mixes the chain rule, product rule and quotient rule, so first decide which one the function needs. Type the derivative, then press Mark to check it in the course.

Proof of the product rule

The product rule comes from differentiation from first principles. Let y = f(x)g(x). By definition:

The trick is to subtract and add f(x + h)g(x) on the top, which changes nothing but lets you split the fraction:

As h tends to 0, f(x + h) tends to f(x), the first fraction tends to g′(x) and the second tends to f′(x). That leaves:

Product rule, quotient rule and chain rule

The three rules are often taught together because most harder functions need more than one of them.

• Product rule: for y = uv, dy/dx = u dv/dx + v du/dx.
• Quotient rule: for y = u/v, dy/dx = (v du/dx − u dv/dx)/v². You can always use the product rule instead by writing u/v as u × v⁻¹.
• Chain rule: for y = f(g(x)), dy/dx = f′(g(x)) × g′(x). It differentiates a function inside another function.

Examples 2 and 5 used the chain rule to find dv/dx inside a product rule question. That combination is the most common one in A-level Maths papers. To check any derivative, type it into the Derivative Calculator, which shows each rule it applies.

Product rule in integration

There is no simple product rule for integration. The reverse of the product rule is integration by parts:

It comes from integrating both sides of the product rule and rearranging, so it is the product rule run backwards.

Common mistakes

• Multiplying the derivatives. The derivative of uv is u′v + uv′, never u′v′.
• Forgetting the chain rule inside. In x² e^(3x), the derivative of e^(3x) is 3e^(3x), not e^(3x).
• Mixing it up with the quotient rule. The product rule has a plus sign and no denominator. The quotient rule has a minus sign, and the order of its terms matters.
• Using it when you don't need to. A constant factor stays in front, and a short polynomial product is quicker to expand.
• Leaving the answer unsimplified. If the question says "show that" or asks for a factorised form, take out common factors at their lowest power.
• Dropping a negative sign. The derivative of cos x is −sin x. A lost minus sign is the most common way to lose the final mark.

Is the product rule in the formula booklet?

No, not for A-level Maths. This was checked against the exam boards' booklets in November 2026.

So learn the product rule by heart. The quotient rule is printed in both booklets, but it is quicker to know it too. Integration by parts is printed, so you do not need to memorise that one, but you do need to choose u and dv/dx well.

Practise differentiation in a short timed paper

A-level Maths: Calculus gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across differentiation and integration, so the product rule comes up often but not in every paper. In the US, the same practice opens as AP Calculus AB.

Read next

Maths · CalculusChain rule: formula, examples and how to use itThe chain rule formula, how to differentiate a function of a function step by step, 8 worked examples, practice questions with answers and exam tips.