Using the Product Rule in Differentiation (With Step-by-Step Examples)

When you are learning calculus for the first time, the product rule is one of the first “real” tools you add to your toolbox.

As a tutor, I often see students try to differentiate a product by just differentiating each factor separately. That does not work in general. The product rule is the fix.

This post walks through:

• What the product rule says

• How to spot when you need it

• One easy, one medium, and one hard example, each with line-by-line solutions

What is the Product Rule?

Suppose

y = f(x)g(x)

is a product of two differentiable functions. The product rule says

y' = f'(x)g(x) + f(x)g'(x)

In words:

1. Differentiate the first function and keep the second as it is.

2. Add the first function times the derivative of the second.

I often teach it as:

• Derivative of the first times the second

• plus the first times the derivative of the second.

How to Recognize When You Need the Product Rule

You need the product rule when:

• Your function is written as one expression multiplied by another, and

• Both parts depend on x.

  
  

Some classic patterns:

• Polynomial × polynomial, e.g. (2x - 1)(x³ + 4)

• Polynomial × trig, e.g. x² ⋅ sin(x)

• Polynomial × exponential, e.g. x ⋅ eˣ

• Log × polynomial, e.g. ln(x) ⋅ x³

  
  

If a factor is a plain constant, you do not need the product rule. You can just pull the constant out.

Easy Example: Polynomial Times Trig

Differentiate

y = x² ⋅ sin(x)

Step 1: Identify the two functions

Let

f(x) = x², g(x) = sin(x)

Step 2: Find each derivative

f'(x) = 2x, g'(x) = cos(x)

Step 3: Apply the product rule

y' = f'(x)g(x) + f(x)g'(x)

Substitute:

y' = (2x)(sin(x)) + (x²)(cos(x))

That is the derivative.

Medium Example: Polynomial Times Polynomial

Differentiate and simplify

y = (3x² − 1)(5x³ + 2x)

You could multiply everything out first, then differentiate. Here we will practice the product rule directly, then simplify the result.

Step 1: Identify the functions

f(x) = (3x² − 1), g(x) = (5x³ + 2x)

Step 2: Differentiate each one

f'(x) = 6x, g'(x) = 15x² + 2

Step 3: Apply the product rule

y' = f'(x)g(x) + f(x)g'(x)

Substitute:

y' = (6x)(5x³ + 2x) + (3x² − 1)(15x² + 2)

Step 4: Expand each product

y' = 30x⁴ + 12x² + 45x⁴ + 6x² − 15x² − 2

Step 5: Combine like terms

y' = 30x⁴ + 12x² + 45x⁴ + 6x² − 15x² − 2

y' = 30x⁴ + 45x⁴ + 12x² + 6x² − 15x² − 2

y' = 75x⁴ + 3x² − 2

That is the simplified derivative.

Hard Example: Product of Three Factors (Using Product Rule + Chain Rule)

Differentiate

y = x² ⋅ eˣ ⋅ ln(3x)

for x > 0.

This is a “triple product”: polynomial × exponential × log. You can handle it by expanding the regular product law as follows:

Suppose

y = f(x)g(x)h(x)

is a product of three differentiable functions. The product rule says

y' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)

In words:

1. Differentiate the first function and keep the second and third as they are.

2. Add the first function times the derivative of the second times the third.

3. Add the first function times the second times the derivative of the third.

I often teach it as:

• Differentiate the first function and keep the second and third as they are.

• Then add the first function times the derivative of the second, keeping the third the same.

• Then add the first and second functions times the derivative of the third.

Let's return to the example

Step 1: Identify the three functions

f(x) = x², g(x) = eˣ, h(x) = ln(3x)

Step 2: Differentiate each one

f'(x) = 2x, g'(x) = eˣ, h'(x) = (1/3x)(3) = (1/x)

Note that we needed the chain rule to differentiate ln(3x).

Step 3: Apply the expanded product rule

y' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)

Substitute:

y' = (2x)(eˣ)(ln(3x)) + (x²)(eˣ)(ln(3x)) + (x²)(eˣ)(1/x)

Step 4: Simplify and factor

y' = (2x)(eˣ)(ln(3x)) + (x²)(eˣ)(ln(3x)) + (x²)(eˣ)(1/x)

y' = (2x)(eˣ)(ln(3x)) + (x²)(eˣ)(ln(3x)) + (x)(eˣ)

y' = (xeˣ)(2ln(3x) + xln(3x) + 1)

That is the derivative.

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