A Level Product Rule – Complete Exam Guide with Worked Examples

Let’s take a look at the following:

y = x^2 e^{3x}

Now, if this appeared in an exam, many students would immediately try to start differentiating. That instinct — to just begin — is usually where small errors creep in. So before we do anything technical, let’s slow down and actually look at what the expression is made of.

What do we have here?

We have a polynomial, x^2, and we have an exponential expression, e^{3x}. And crucially, they are being multiplied together.

That multiplication is the important part. Whenever two different functions are multiplied, that is your signal that product rule is involved. It doesn’t matter how simple the functions look individually — the multiplication forces the structure.

So rather than thinking “differentiate the whole thing”, we train ourselves to think: this is one function multiplied by another.

That shift in mindset is what protects marks.

Separating the Two Parts

To keep things controlled, I’m going to give each part a name. Not because the letters matter, but because it forces us to treat them independently.

Let
u = x^2

and
v = e^{3x}

The product rule tells us that:

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Now pause for a moment and think about what that rule is really saying.

It’s saying: differentiate one part while keeping the other one exactly as it is. Then swap them over.

That’s it.

There will always be two terms at the end. If you only see one, you know something has gone wrong.

Differentiating Each Piece

Let’s deal with them one at a time.

First, the polynomial part. Differentiating x^2 gives:

\frac{du}{dx} = 2x

Nothing surprising there.

Now we move to the exponential part, e^{3x}. This is where students sometimes hesitate, because it’s not just “e to the x”. There’s a 3x inside.

Whenever you differentiate something of the form e^{f(x)}, the exponential stays exactly the same, but you multiply by the derivative of what’s inside.

So differentiating e^{3x} gives:

\frac{dv}{dx} = 3e^{3x}

That 3 appears because the derivative of 3x is 3.

This is chain rule quietly operating inside the product rule. And that combination is very common at A Level — examiners like layering rules together.

Putting It Together

Now we apply the product rule formula carefully.

Remember: first function multiplied by the derivative of the second, plus second function multiplied by the derivative of the first.

So we write:

\frac{dy}{dx} = x^2(3e^{3x}) + e^{3x}(2x)

At this stage, the method is complete. Even if the algebra hasn’t been tidied, the structure is correct. In an exam, that structure earns the method marks.

Students often rush here and start expanding brackets unnecessarily. There’s no need. In fact, expanding too early often creates avoidable algebra errors.

Simplifying Sensibly

Both terms contain e^{3x}, so it makes sense to factor it out. Doing this makes the expression cleaner and easier to work with later — especially if you were solving for stationary points.

So we factor:

\frac{dy}{dx} = e^{3x}(3x^2 + 2x)

You could factor further if you wanted:

\frac{dy}{dx} = xe^{3x}(3x + 2)

Either form is perfectly acceptable unless the question specifies otherwise.

What I’d Say in a Classroom

If I were teaching this live, I’d emphasise that the difficulty here isn’t the differentiation itself. It’s recognising the structure before you start.

Most lost marks in product rule questions come from one of three things:

• forgetting the second term entirely
  • missing the inner derivative in the exponential
  • rushing simplification and creating algebra mistakes

None of those are “knowledge” problems. They’re structure problems.

When you slow down at the beginning and ask yourself, “What type of structure am I looking at?”, the rest becomes procedural. Calm. Predictable.

And that’s exactly what you want differentiation to feel like in an exam.