Product, quotient and chain rules in combination (QCE Mathematical Methods Unit 3)

Syllabus dot point

“Apply the product, quotient and chain rules, including in combination, to differentiate functions built from polynomial, exponential, logarithmic and trigonometric components”

8 min readUpdated 7 Oct 2026

Quick answer

The product rule gives $(uv)' = u'v + uv'$ , the quotient rule gives $(u/v)' = (u'v - uv')/v^2$ , and the chain rule gives $(f(g(x)))' = f'(g(x)) g'(x)$ , and they combine in a clear inside-to-outside order whenever a function is built from polynomial, exponential, logarithmic and trigonometric pieces.

Jump to a section

What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to differentiate any function built from the standard library (polynomials, $e^x$ , $\ln x$ , $\sin x$ , $\cos x$ , $\tan x$ ) using combinations of the product, quotient and chain rules. Almost every Methods calculus question starts with a derivative step, so fluency here is non-negotiable for IA2, Paper 1 of the EA, and the calculus content of every PSMT.

The answer

The chain rule

If $y = f(g(x))$ , set $u = g(x)$ so $y = f(u)$ . Then

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

In practice: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.

The product rule

If $y = u(x) \, v(x)$ , then

\frac{dy}{dx} = u' v + u v'.

The quotient rule

If $y = \dfrac{u(x)}{v(x)}$ with $v(x) \neq 0$ , then

\frac{dy}{dx} = \frac{u' v - u v'}{v^2}.

The numerator is $u' v$ minus $u v'$ , in that order. Reversing the order changes the sign.

Standard derivatives (the library)

The rules above act on the standard derivatives. Memorise these.

\frac{d}{dx}(x^n) = n x^{n - 1} \quad (\text{any real } n)

\frac{d}{dx}(e^x) = e^x \qquad \frac{d}{dx}(\ln x) = \frac{1}{x}

\frac{d}{dx}(\sin x) = \cos x \qquad \frac{d}{dx}(\cos x) = -\sin x \qquad \frac{d}{dx}(\tan x) = \sec^2 x

Order of operations

When two or more rules combine, choose the outer structure first and work inwards.

$y = (x^2 + 1) \sin x$ is a product; differentiate with the product rule, and the inside of $\sin x$ is just $x$ so no chain rule on the trig.
$y = \sin(x^2 + 1)$ is a composition; differentiate with the chain rule, no product rule.
$y = \sin(x^2 + 1) \cdot e^{3x}$ is a product whose factors each need the chain rule; apply the product rule and the chain rule lives inside each derivative.

Choosing the rule from the structure

The first decision is to read the outermost structure of the expression. If the whole thing is two factors multiplied, it is a product; if it is one expression divided by another, it is a quotient; if it is one function applied to another, it is a composition needing the chain rule. Only after fixing the outer rule do you differentiate the inner parts, which may themselves need further rules. Mislabelling the outer structure (treating a composition as a product, say) is the most common source of error, so naming it explicitly before differentiating is worth the moment it takes.

Simplifying before and after

Some derivatives are far easier after a preliminary simplification. A quotient of powers such as $\dfrac{x^3 - 2x}{x}$ should be split into $x^2 - 2$ before differentiating, avoiding the quotient rule entirely. After differentiating a product or quotient, look for a common factor to extract: QCAA frequently allocates a mark for a clean factored final form, and an unsimplified expression can also hide sign errors. The habit of simplifying at both ends turns long calculations into short ones.

Worked example

Five differentiation problems combining the product, quotient and chain rules in various arrangements.

Step 1: Differentiate $y = x \sin(2x)$ using the product rule with a chain inside

The outer structure is a product of $u = x$ and $v = \sin(2x)$ , so we apply the product rule. To find $v'$ , we need the chain rule on $\sin(2x)$ : the inner function $2x$ gives a factor of $2$ .

u' = 1, \quad v' = 2\cos(2x)

\frac{dy}{dx} = u'v + uv' = \sin(2x) + 2x\cos(2x)

Step 2: Differentiate $y = \dfrac{e^{2x}}{x + 1}$ using the quotient rule with a chain inside

The outer structure is a quotient, so we apply the quotient rule. To find $u'$ , the chain rule on $e^{2x}$ gives a factor of $2$ .

u = e^{2x},\; v = x + 1,\; u' = 2e^{2x},\; v' = 1

\frac{dy}{dx} = \frac{2e^{2x}(x + 1) - e^{2x} \cdot 1}{(x + 1)^2}

Factoring $e^{2x}$ from the numerator simplifies the answer.

\frac{dy}{dx} = \frac{e^{2x}(2x + 1)}{(x + 1)^2}

Step 3: Apply a double chain to $y = \sin^2(3x) = (\sin(3x))^2$

There are three nested layers: the outer squaring, then $\sin$ , then $3x$ . We differentiate from the outside inwards, collecting a factor from each layer.

\frac{dy}{dx} = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x) = 3\sin(6x)

The last step uses the double-angle identity $2\sin\theta\cos\theta = \sin(2\theta)$ .

Step 4: Differentiate $y = x \ln x$ using the product rule

The outer structure is a product of $u = x$ and $v = \ln x$ . The derivative of $\ln x$ is $\tfrac{1}{x}$ , so the $x$ in $u\,v'$ cancels neatly.

\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1

Step 5: Differentiate $y = \dfrac{\cos(3x)}{x^2}$ using the quotient rule with a chain inside

The outer structure is a quotient. To find $u'$ , the chain rule on $\cos(3x)$ contributes a factor of $3$ and changes the sign.

u = \cos(3x),\; v = x^2,\; u' = -3\sin(3x),\; v' = 2x

\frac{dy}{dx} = \frac{-3\sin(3x) \cdot x^2 - \cos(3x) \cdot 2x}{x^4} = \frac{-3x\sin(3x) - 2\cos(3x)}{x^3}

Final answer: the common factor of $x$ has been cancelled from the numerator and denominator to reach the simplified form.

Common traps

Forgetting the chain rule on composed functions: Writing $\frac{d}{dx} \sin(2x) = \cos(2x)$ drops the factor of $2$ . The correct answer is $2 \cos(2x)$ .
Reversing the quotient rule sign: The numerator is $u' v$ minus $u v'$ . Writing it the other way around flips the sign of the entire derivative.
Treating $e^{2x}$ like a power: $\frac{d}{dx}(e^{2x}) = 2 e^{2x}$ , by the chain rule. It is not $2x \, e^{2x - 1}$ .
Applying the power rule to $a^x$: For non- $e$ exponentials, $\frac{d}{dx}(a^x) = (\ln a) \, a^x$ . The power rule does not apply because the base is constant and the exponent is variable.
Not simplifying: QCAA frequently allocates a mark for a clean final form. After a product or quotient rule, look for common factors and factor them out.
Choosing the wrong outer structure: $y = e^{x^2 + x}$ is a composition, not a product. Apply the chain rule, not the product rule.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Paper 1 (technique). Differentiate

y = x^2 e^{3x}

with respect to

x

, and write your answer in factored form.

Show worked answer →

Use the product rule with $u = x^2$ and $v = e^{3x}$ .

$u' = 2x$ , $v' = 3 e^{3x}$ (chain rule on the inner $3x$ ).

$\frac{dy}{dx} = u' v + u v' = 2x \, e^{3x} + 3 x^2 e^{3x}.$

Factor: $\frac{dy}{dx} = x \, e^{3x} (2 + 3x).$

Markers reward the explicit product rule setup, both derivatives correct (including the chain rule factor on $e^{3x}$ ), and the factored final answer. QCAA typically awards the final mark for clean factorisation.

Practice3 marks

Paper 1 (technique). Determine

\frac{dy}{dx}

for

y = \frac{\ln x}{x^2}

Show worked answer →

Quotient rule with $u = \ln x$ and $v = x^2$ .

$u' = \frac{1}{x}$ , $v' = 2x$ .

$\frac{dy}{dx} = \frac{u' v - u v'}{v^2} = \frac{\frac{1}{x} \cdot x^2 - \ln x \cdot 2x}{x^4} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2 \ln x}{x^3}.$

Markers reward the correct quotient rule order ( $u' v$ minus $u v'$ ), accurate derivatives, and simplification by cancelling the common factor of $x$ .

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA