What is wrong numerator order in the quotient rule?

It is u'v-uv', not uv'-u'v. Reversing the order negates the whole derivative.

What are sign slips with trig and exponential factors?

Track the minus from ddxcos x=-sin x carefully when it meets the minus in the quotient rule.

SCSA WACE 2021-style practice: Calculator-free. Differentiate with respect to x: (a) y=x^3ln x; (b) y=e^xx+1.

Calculator-free practice of both rules. (a) Product rule with u=x^3, v=ln x: u'=3x^2, v'=1x. So dydx=3x^2ln x+x^3·1x=3x^2ln x+x^2=x^2(3ln x+1). (b) Quotient rule with u=e^x, v=x+1: u'=e^x, v'=1. So dydx=e^x(x+1)-e^x(1)(x+1)^2=e^x\,x(x+1)^2=xe^x(x+1)^2. Markers reward correct u,v choices, the internal derivatives, and a factored final form.

SCSA WACE 2023-style practice: Calculator-assumed. For f(x)=xe^x, find f'(x) and hence the coordinates of the stationary point.

A quotient-rule application leading to a stationary point. Quotient rule with u=x, v=e^x: u'=1, v'=e^x. So f'(x)=1· e^x-x e^x(e^x)^2=e^x(1-x)e^2x=1-xe^x. Stationary point. f'(x)=0 when 1-x=0, so x=1. Then f(1)=1e≈ 0.368. The stationary point is (1,1e). Markers reward the quotient rule, cancelling e^x, and the correct stationary point. Writing y=xe^-x and using the product rule is an equally valid route.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we differentiate a function that is a product or a quotient of two simpler functions?

Establish and apply the product rule and the quotient rule to differentiate products and quotients of functions

WACE Year 12 Mathematics Methods Unit 3 the product and quotient rules: their statements, when to use each, applying them to polynomial, exponential and trigonometric factors, with worked SCSA-style examples and the numerator order trap.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The product rule
The quotient rule
Combining the rules with the chain rule
Choosing between rewriting and the quotient rule
Three or more factors

What this dot point is asking

SCSA Unit 3 establishes two of the three combining rules for differentiation. This dot point asks you to recognise product and quotient structures and apply the correct rule, which is examined heavily in the calculator-free section where by-hand fluency is tested.

The product rule

When a function is a product of two factors that each depend on $x$ , you cannot simply multiply the derivatives. The product rule accounts for both factors changing.

The quotient rule

When a function is one expression divided by another, the quotient rule applies. The numerator order matters: the derivative of the top times the bottom comes first.

Applying the quotient rule

Step 1: Identify numerator and denominator, then differentiate each

The function $y = \dfrac{x}{\cos x}$ is a quotient, so the quotient rule applies. Label $u=x$ and $v=\cos x$ . Differentiating gives $u'=1$ and $v'=-\sin x$ .

Step 2: Substitute into $y'=\dfrac{u'v-uv'}{v^{2}}$ and simplify

The minus in the quotient rule numerator and the minus from $v'=-\sin x$ combine to give a positive term, which is the most common source of sign errors:

\frac{dy}{dx} = \frac{(1)(\cos x) - (x)(-\sin x)}{\cos^{2}x} = \frac{\cos x + x\sin x}{\cos^{2}x}.

Final answer: $\dfrac{dy}{dx} = \dfrac{\cos x + x\sin x}{\cos^{2}x}$

Combining the rules with the chain rule

In practice the product and quotient rules rarely appear alone: one or both factors is itself a composite needing the chain rule. The method is to differentiate each factor fully (chain rule included) before substituting into the product or quotient rule.

A product where each factor needs the chain rule

Step 1: Label the factors and apply the chain rule to each

The function $y = e^{2x}\sin(3x)$ is a product where both factors are composites, so the chain rule must be applied before the product rule. Label $u=e^{2x}$ and $v=\sin(3x)$ . The chain rule gives $u'=2e^{2x}$ (inner derivative $2$ ) and $v'=3\cos(3x)$ (inner derivative $3$ ).

Step 2: Apply $y'=u'v+uv'$ and factorise the common $e^{2x}$

Substituting into the product rule and factoring the common factor produces the compact final form:

\frac{dy}{dx}=2e^{2x}\sin(3x)+e^{2x}\cdot 3\cos(3x)=e^{2x}\bigl(2\sin(3x)+3\cos(3x)\bigr).

Final answer: $\dfrac{dy}{dx} = e^{2x}\bigl(2\sin(3x)+3\cos(3x)\bigr)$

Choosing between rewriting and the quotient rule

A quotient can sometimes be rewritten as a product with a negative power, then differentiated with the product or chain rule. For instance $\dfrac{x}{\cos x}=x\sec x$ or $\dfrac{1}{x^{2}}=x^{-2}$ . Choose whichever path is cleaner; for a simple power in the denominator, rewriting is usually faster than the quotient rule.

Three or more factors

The product rule extends to a product of three functions by differentiating each in turn while holding the others fixed: $\dfrac{d}{dx}(uvw)=u'vw+uv'w+uvw'$ . In practice it is usually cleaner to group two factors as a single $U=uv$ , apply the two-factor rule to $Uw$ , and expand. SCSA questions rarely demand three factors directly, but recognising the pattern avoids errors when a product is hidden inside a larger expression.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20215 marksCalculator-free. Differentiate with respect to

x

: (a)

y=x^{3}\ln x

; (b)

y=\dfrac{e^{x}}{x+1}

Show worked answer →

Calculator-free practice of both rules.

(a) Product rule with $u=x^{3}$ , $v=\ln x$ : $u'=3x^{2}$ , $v'=\dfrac{1}{x}$ . So $\dfrac{dy}{dx}=3x^{2}\ln x+x^{3}\cdot\dfrac{1}{x}=3x^{2}\ln x+x^{2}=x^{2}(3\ln x+1)$ .

(b) Quotient rule with $u=e^{x}$ , $v=x+1$ : $u'=e^{x}$ , $v'=1$ . So $\dfrac{dy}{dx}=\dfrac{e^{x}(x+1)-e^{x}(1)}{(x+1)^{2}}=\dfrac{e^{x}\,x}{(x+1)^{2}}=\dfrac{xe^{x}}{(x+1)^{2}}$ .

Markers reward correct $u,v$ choices, the internal derivatives, and a factored final form.

WACE 20234 marksCalculator-assumed. For

f(x)=\dfrac{x}{e^{x}}

, find

f'(x)

and hence the coordinates of the stationary point.

Show worked answer →

A quotient-rule application leading to a stationary point.

Quotient rule with $u=x$ , $v=e^{x}$ : $u'=1$ , $v'=e^{x}$ . So $f'(x)=\dfrac{1\cdot e^{x}-x e^{x}}{(e^{x})^{2}}=\dfrac{e^{x}(1-x)}{e^{2x}}=\dfrac{1-x}{e^{x}}$ .

Stationary point. $f'(x)=0$ when $1-x=0$ , so $x=1$ . Then $f(1)=\dfrac{1}{e}\approx 0.368$ . The stationary point is $\left(1,\dfrac{1}{e}\right)$ .

Markers reward the quotient rule, cancelling $e^{x}$ , and the correct stationary point. Writing $y=xe^{-x}$ and using the product rule is an equally valid route.