The Product Rule

Differentiating products of two functions

SPEC 7.4 • DIFFERENTIATION

Why We Need the Product Rule

When we have a function that is a product of two simpler functions, we cannot just differentiate each part separately.

Common Mistake

Consider

y = x \times x^{2} = x^{3}

We know

\frac{d y}{d x} = 3 x^{2}

But if we differentiated each part separately:

1 \times 2 x = 2 x

✗

This is WRONG!

Key Insight

When both parts of a product contain the variable

x

, we need a special rule to differentiate correctly.

The Product Rule

If

y = uv

where

u

and

v

are both functions of

x

, then:

\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}

Memory Aid

"First times derivative of second, plus second times derivative of first"

Or in shorthand:

(uv)^{'} = u v^{'} + v u^{'}

Note

It doesn't matter which function you call

u

and which you call

v

— the result will be the same!

Systematic Approach

Follow these steps every time you use the product rule:

Method

StepAction1Identify

u

and

v

2Find

\frac{d u}{d x}

3Find

\frac{d v}{d x}

4Apply:

\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}

5Simplify and factorise if possible

Exam Tip

Writing out

u

,

v

,

\frac{d u}{d x}

,

\frac{d v}{d x}

clearly before substituting helps avoid errors and earns method marks.

Worked Example01

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

1

Let

u = x^{3}

and

v = 2 x - 5

2

\frac{d u}{d x} = 3 x^{2}

3

\frac{d v}{d x} = 2

4

\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x} = x^{3} (2) + (2 x - 5) (3 x^{2})

5

\frac{d y}{d x} = 2 x^{3} + 6 x^{3} - 15 x^{2} = 8 x^{3} - 15 x^{2}

Answer: $\frac{d y}{d x} = 8 x^{3} - 15 x^{2}$ or $x^{2} (8 x - 15)$

Worked Example02

Differentiate $y = x^{2} e^{x}$

1

Let

u = x^{2}

and

v = e^{x}

2

\frac{d u}{d x} = 2 x

3

\frac{d v}{d x} = e^{x}

4

\frac{d y}{d x} = x^{2} (e^{x}) + e^{x} (2 x)

5

\frac{d y}{d x} = x^{2} e^{x} + 2 x e^{x} = e^{x} (x^{2} + 2 x)

Answer: $\frac{d y}{d x} = e^{x} (x^{2} + 2 x)$ or $x e^{x} (x + 2)$

Worked Example03

Differentiate $y = x sin (x)$

1

Let

u = x

and

v = sin (x)

2

\frac{d u}{d x} = 1

3

\frac{d v}{d x} = cos (x)

4

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

Answer: $\frac{d y}{d x} = x cos (x) + sin (x)$

Worked Example04

Differentiate $y = x^{3} ln (x)$

1

Let

u = x^{3}

and

v = ln (x)

2

\frac{d u}{d x} = 3 x^{2}

3

\frac{d v}{d x} = \frac{1}{x}

4

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

5

\frac{d y}{d x} = x^{2} + 3 x^{2} ln (x) = x^{2} (1 + 3 ln (x))

Answer: $\frac{d y}{d x} = x^{2} (1 + 3 ln (x))$

Product Rule with Chain Rule

When

u

or

v

is a composite function, you need to use the chain rule when finding

\frac{d u}{d x}

or

\frac{d v}{d x}

.

Example Structure

For

y = (3 x + 2)^{4} \cdot e^{2 x}

:

u = (3 x + 2)^{4}

→

\frac{d u}{d x} = 4 (3 x + 2)^{3} \times 3 = 12 (3 x + 2)^{3}

(chain rule!)

v = e^{2 x}

→

\frac{d v}{d x} = 2 e^{2 x}

(chain rule!)

Strategy

Find each derivative separately and carefully, then combine using the product rule formula.

Worked Example05

Differentiate $y = (3 x + 2)^{4} \cdot e^{2 x}$

1

u = (3 x + 2)^{4}

,

v = e^{2 x}

2

\frac{d u}{d x} = 4 (3 x + 2)^{3} \times 3 = 12 (3 x + 2)^{3}

3

\frac{d v}{d x} = 2 e^{2 x}

4

\frac{d y}{d x} = (3 x + 2)^{4} (2 e^{2 x}) + e^{2 x} (12 (3 x + 2)^{3})

5

Factor out

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

:

•

= 2 e^{2 x} (3 x + 2)^{3} [(3 x + 2) + 6]

Answer: $\frac{d y}{d x} = 2 e^{2 x} (3 x + 2)^{3} (3 x + 8)$

Worked Example06

Differentiate $y = sin (2 x) \cdot cos (3 x)$

1

u = sin (2 x)

,

v = cos (3 x)

2

\frac{d u}{d x} = 2 cos (2 x)

(chain rule)

3

\frac{d v}{d x} = - 3 sin (3 x)

(chain rule)

4

\frac{d y}{d x} = sin (2 x) (- 3 sin (3 x)) + cos (3 x) (2 cos (2 x))

Answer: $\frac{d y}{d x} = 2 cos (2 x) cos (3 x) - 3 sin (2 x) sin (3 x)$

Worked Example07

Differentiate $y = cos (2 x) \cdot ln (3 x)$

1

u = cos (2 x)

,

v = ln (3 x)

2

\frac{d u}{d x} = - 2 sin (2 x)

(chain rule)

3

\frac{d v}{d x} = \frac{1}{3 x} \times 3 = \frac{1}{x}

(chain rule)

4

\frac{d y}{d x} = cos (2 x) \cdot \frac{1}{x} + ln (3 x) \cdot (- 2 sin (2 x))

5

\frac{d y}{d x} = \frac{c o s ( 2 x )}{x} - 2 sin (2 x) ln (3 x)

Answer: $\frac{d y}{d x} = \frac{c o s ( 2 x )}{x} - 2 sin (2 x) ln (3 x)$

Worked Example08

Differentiate $y = e^{- x} \cdot ln (x^{2} + 1)$

1

u = e^{- x}

,

v = ln (x^{2} + 1)

2

\frac{d u}{d x} = - e^{- x}

(chain rule)

3

\frac{d v}{d x} = \frac{1}{x ^{2} + 1} \times 2 x = \frac{2 x}{x ^{2} + 1}

(chain rule)

4

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

5

Factor out

e^{- x}

:

Answer: $\frac{d y}{d x} = e^{- x} (\frac{2 x}{x ^{2} + 1} - ln (x^{2} + 1))$

Practice

Question 1: Differentiate

y = (x^{2} + 1) \cdot sin (2 x)

1

u = x^{2} + 1

,

v = sin (2 x)

2

\frac{d u}{d x} = 2 x

,

\frac{d v}{d x} = 2 cos (2 x)

3

\frac{d y}{d x} = (x^{2} + 1) (2 cos 2 x) + sin (2 x) (2 x)

Answer: $\frac{d y}{d x} = 2 (x^{2} + 1) cos (2 x) + 2 x sin (2 x)$

Practice

Question 2: Differentiate

y = e^{3 x} \cdot ln (2 x + 1)

1

u = e^{3 x}

,

v = ln (2 x + 1)

2

\frac{d u}{d x} = 3 e^{3 x}

,

\frac{d v}{d x} = \frac{2}{2 x + 1}

3

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

4

Factor out

e^{3 x}

:

Answer: $\frac{d y}{d x} = e^{3 x} (\frac{2}{2 x + 1} + 3 ln (2 x + 1))$

Practice

Question 3: Differentiate

y = sin (x) \cdot ln (x)

1

u = sin (x)

,

v = ln (x)

2

\frac{d u}{d x} = cos (x)

,

\frac{d v}{d x} = \frac{1}{x}

3

\frac{d y}{d x} = sin (x) \cdot \frac{1}{x} + ln (x) \cdot cos (x)

Answer: $\frac{d y}{d x} = \frac{s i n ( x )}{x} + cos (x) ln (x)$

Exam Tips

For Maximum Marks

Write out

u

,

v

,

\frac{d u}{d x}

,

\frac{d v}{d x}

clearly — this earns method marks even if you make an error later

State the product rule formula if asked to "show that" or "prove"

Always factorise your final answer where possible

Apply chain rule carefully when

u

or

v

is a composite function

Check by expanding — for simple products, you can verify by expanding first then differentiating

Recognition

Learn to recognise when a function is a product of two parts — this is the first step!

Summary

Key Concepts

ConceptFormula/Method

Product Rule

\frac{d}{d x} [uv] = u \frac{d v}{d x} + v \frac{d u}{d x}

Alternative form

(uv)^{'} = u v^{'} + v u^{'}

With chain ruleApply chain rule when finding

u^{'}

or

v^{'}

if composite

FactorisingAlways look to factor out common terms in final answer

Remember

The product rule is essential for differentiating functions like

x^{n} e^{x}

,

x sin x

,

x ln x

, and any other product of two functions of

x

.