G4PureStretch

The Chain, Product and Quotient Rules

Q: Is the quotient rule numerator order important?

Yes — it is NOT symmetric. It must be v dudx-u dvdx (the derivative of the top comes first, then subtract), all over v^2.

The power rule alone cannot differentiate $(3 x + 1)^{5}$ , $x^{2} sin x$ or $\frac{x}{x + 1}$ — none of these is a simple power of $x$ . Three rules extend differentiation to every combination of functions you will meet: composite functions (chain), products (product), and quotients (quotient).

36 min Video by TLMaths Differentiation

The big picture

Almost every "interesting" function in mathematics is built by combining simpler functions — nesting them, multiplying them, or dividing them. The chain rule differentiates a function INSIDE another function; the product rule differentiates two functions MULTIPLIED together; the quotient rule differentiates one function DIVIDED by another. Mastering all three turns differentiation from "a trick for powers of $x$ " into a tool that works on almost anything.

A-Level Maths G4-01: Introducing the Chain RuleWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Apply the chain rule to differentiate composite functions
Apply the product rule to differentiate a product of two functions
Apply the quotient rule to differentiate one function divided by another
Choose the correct rule (or combination of rules) for a given expression

The chain rule

When a function is "inside" another — like $(3 x + 1)^{5}$ , where $3 x + 1$ is inside the power of 5 — differentiate the OUTER function first (treating the inner function as a single variable), then multiply by the derivative of the INNER function.

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}, where u is the inner function

Differentiate the outer function, multiply by the derivative of the inner.

1Let

u = 3 x + 1

(inner function), so

y = u^{5}

\frac{d y}{d u} = 5 u^{4}

and

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

3Multiply:

\frac{d y}{d x} = 5 u^{4} \times 3 = 15 u^{4}

4Substitute back:

15 (3 x + 1)^{4}

Answer

\frac{d y}{d x} = 15 (3 x + 1)^{4}

Tip — In practice: differentiate the outside "as if the bracket were just x", then multiply by the derivative of what is inside the bracket.

The product rule

When two functions of $x$ are multiplied together, like $x^{2} sin x$ , the product rule says: differentiate the first, keep the second; PLUS keep the first, differentiate the second.

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

u and v are both functions of x.

1Let

u = x^{2}

v = (2 x + 1)^{3}

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

; using the chain rule,

\frac{d v}{d x} = 3 (2 x + 1)^{2} \times 2 = 6 (2 x + 1)^{2}

3Apply the product rule:

\frac{d y}{d x} = x^{2} \times 6 (2 x + 1)^{2} + (2 x + 1)^{3} \times 2 x

4Factorise out common terms

2 x (2 x + 1)^{2}

2 x (2 x + 1)^{2} [3 x + (2 x + 1)] = 2 x (2 x + 1)^{2} (5 x + 1)

Answer

\frac{d y}{d x} = 2 x (2 x + 1)^{2} (5 x + 1)

Why this works

Factorising the final answer (pulling out the lowest powers common to both terms) is often expected in exams — it shows the result in its most compact, recognisable form.

The quotient rule

When one function is divided by another, like $\frac{x}{2 x + 1}$ , use the quotient rule — remember the order in the numerator matters (it is NOT symmetric).

y = \frac{u}{v} \Rightarrow \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

"Low d-high minus high d-low, over low squared".

1Let

u = x

v = 2 x + 1

, so

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

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

2Apply the quotient rule:

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

3Simplify the numerator:

2 x + 1 - 2 x = 1

Answer

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

Tip — A useful mnemonic: "low d-high minus high d-low, over low squared" — where "low" is the denominator $v$ and "high" is the numerator $u$ .

Think like an examiner

Identify which rule applies BEFORE differentiating: is it one function nested in another (chain), two functions multiplied (product), or one divided by another (quotient)?
For the quotient rule, the order in the numerator is critical — $u \frac{d v}{d x} - v \frac{d u}{d x}$ is the wrong way round and a very common error.
Always attempt to factorise or simplify your final answer — unsimplified expressions can lose marks even if the method is correct.
Many questions combine rules (e.g. the chain rule used to differentiate $v$ inside a product rule question) — expect to nest them.

Common misconceptions

Forgetting to multiply by the derivative of the inner function in the chain rule.

You must ALWAYS multiply by

\frac{d u}{d x}

— differentiating only the outer function is an incomplete chain rule.

Getting the numerator order wrong in the quotient rule.

It is

v \frac{d u}{d x} - u \frac{d v}{d x}

— the term with

\frac{d u}{d x}

comes first, with a MINUS sign, not a plus.

Adding instead of the product rule structure, e.g. writing

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

The product rule is

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

— each original function stays attached to the OTHER function's derivative.

Differentiation rules

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

Chain rule.

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

Product rule (y = uv).

\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

Quotient rule (y = u/v).

Stretch yourself

Differentiate $y = \frac{x ^{2}}{( 3 x - 1 ) ^{2}}$ using the quotient rule, giving your answer as a single fraction.

Hint — Let $u = x^{2}$ and $v = (3 x - 1)^{2}$ ; you will need the chain rule to differentiate $v$ .

Questions students ask

How do I know which rule to use?

Ask what the LAST operation is if you were evaluating the expression: if it is "raise a bracket to a power" or similar nesting, use the chain rule; if the final operation is multiplication of two separate functions of

x

, use the product rule; if it is division, use the quotient rule.

Can the rules be combined?

Yes, frequently — for example the product or quotient rule often needs the chain rule just to differentiate one of the two pieces (

u

v

) if that piece is itself a composite function.

Is the quotient rule numerator order important?

Yes — it is NOT symmetric. It must be

v \frac{d u}{d x} - u \frac{d v}{d x}

(the derivative of the top comes first, then subtract), all over

v^{2}

Key takeaways

Chain rule: differentiate the outer function, multiply by the derivative of the inner.
Product rule: $u \frac{d v}{d x} + v \frac{d u}{d x}$ for $y = uv$ .
Quotient rule: $\frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ for $y = \frac{u}{v}$ — order matters.
Rules often combine: the chain rule is frequently needed inside a product or quotient rule calculation.

How this fits the course

Build on

Stationary Points

Implicit and Parametric Differentiation

Leads to

Implicit and Parametric Differentiation

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The Chain, Product and Quotient Rules

The power rule alone cannot differentiate

(3 x + 1)^{5}

x^{2} sin x

\frac{x}{x + 1}

— none of these is a simple power of

x

. Three rules extend differentiation to every combination of functions you will meet: composite functions (chain), products (product), and quotients (quotient).

36 min Video by TLMaths Differentiation