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hscscience Maths Adv · Y11
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Module 3 · L06 of 15 ~35 min +95 XP available

Product & Quotient Rules

When functions are multiplied or divided, the derivative is not simply the product or quotient of the derivatives. A company's profit depends on both price and quantity sold: changing one affects the other. Two elegant rules handle all of this.

Today's hook, If the area of a rectangle is $A = lw$, and both length $l$ and width $w$ are changing over time, how fast is the area changing? It is tempting to say $\frac{dA}{dt} = \frac{dl}{dt} \cdot \frac{dw}{dt}$, but that is wrong. Why?
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Orient and recall

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
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Recall, your gut answer first
+5 XP warm-up

Before we start, what do you already know about this topic?

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The two moves

Work through the core explanation before applying it.

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The two moves
+5 XP to read

There are only two moves in this entire lesson. Lock them into muscle memory and the rest is just calculation.

Move 1: Identify the structure. Decide if the function is a product $uv$, a quotient $u/v$, or can be simplified first before applying a rule.

Move 2: Write out $u$, $v$, $u'$, $v'$ before substituting. Rushing straight to the formula without labelling the parts is the biggest source of errors.

The product rule: y = u·v differentiates to the sum of two cross-terms u′v + uv′.
$$\frac{d}{dx}(uv) = u'v + uv' \qquad \frac{d}{dx}\!\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$
Product rule
$\frac{d}{dx}(uv) = u'v + uv'$. Two cross-terms added together. Neither term can be skipped.
Quotient rule
$\frac{u'v - uv'}{v^2}$. Note the minus sign and $v^2$ denominator, both are frequently forgotten.
Simplify first?
For $y = x^2 \cdot x^3$, write $y = x^5$ and use the power rule, faster than product rule every time.
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What you will master

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

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What you will master
Know

Key facts

  • The product rule: $(uv)' = u'v + uv'$
  • The quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
  • When simplification is faster than either rule
Understand

Concepts

  • Why the derivative of a product is not the product of the derivatives
  • How both functions contribute cross-terms in the product rule
  • The role of $v^2$ in the quotient rule denominator
Can do

Skills

  • Apply the product rule to differentiate products of two functions
  • Apply the quotient rule to differentiate quotients of two functions
  • Differentiate combinations involving product, quotient, and chain rules
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Key terms
Product ruleIf $y = uv$, then $y' = u'v + uv'$ where $u$ and $v$ are functions of $x$.
Quotient ruleIf $y = \frac{u}{v}$, then $y' = \frac{u'v - uv'}{v^2}$.
Differentiation by simplificationRewriting a product or quotient as a single power or sum before differentiating.
Cross-termsThe two terms $u'v$ and $uv'$ that arise from the product rule, capturing how each factor changes.
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The product and quotient rules

Work through the core explanation before applying it.

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The product and quotient rules
core concept

When two functions are multiplied, the rate of change of the product depends on how each function changes and the value of the other. Both contribute simultaneously, the product rule captures this interaction with two cross-terms:

$$\frac{d}{dx}(uv) = u'v + uv'$$

For quotients, the rate depends on the relative rates of numerator and denominator, weighted by $v^2$ to account for the shrinking or stretching effect of division:

$$\frac{d}{dx}\!\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$
The rectangle analogy. Imagine a rectangle with side lengths $u$ and $v$. As both change, the new area is approximately $(u + \Delta u)(v + \Delta v) = uv + u\,\Delta v + v\,\Delta u + \Delta u\,\Delta v$. The last term is negligibly small, leaving area change $\approx u\,\Delta v + v\,\Delta u$. Dividing by $\Delta x$ and taking the limit gives $\frac{dA}{dx} = uv' + vu'$, the product rule.

Product rule: if $y = uv$, then $y' = u'v + uv'$; Quotient rule: if $y = \frac{u}{v}$, then $y' = \frac{u'v - uv'}{v^2}$

Pause, copy both rules: product rule $y' = u'v + uv'$ and quotient rule $y' = \dfrac{u'v - uv'}{v^2}$ (note the minus sign and the $v^2$ denominator) into your book.

Quick check: True or false, the derivative of a product $uv$ is equal to the product of the derivatives $u'v'$.

Worked examples · 3 in a row, reveal as you go

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Work examples end to end

Follow the reasoning through complete worked solutions.

PROBLEM 1 · PRODUCT RULE WITH $e^x$

Differentiate $y = x^2 e^x$ using the product rule.

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Let $u = x^2$ and $v = e^x$
Identify the two factors of the product.
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PROBLEM 2 · QUOTIENT RULE WITH POLYNOMIAL

Differentiate $y = \dfrac{x^3}{x + 1}$ using the quotient rule.

ApplyBand 4
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Let $u = x^3$ and $v = x + 1$
Identify numerator $u$ and denominator $v$.
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PROBLEM 3 · QUOTIENT RULE WITH $e^x$

Differentiate $y = \dfrac{e^x}{x^2 + 1}$.

ApplyBand 5
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Let $u = e^x$ and $v = x^2 + 1$
Set up for quotient rule with an exponential numerator.
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Quick check: Differentiating $y = xe^x$ using the product rule gives:

Common errors · the 3 traps that cost marks

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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

Trap 01
Using the product rule when you should simplify first
For $y = x^2 \cdot x^3$, it is faster to write $y = x^5$ and use the power rule. Always check if simplification is possible before choosing a rule.
Trap 02
Forgetting to square the denominator in the quotient rule
The quotient rule has $v^2$ in the denominator. Students sometimes write just $v$ or forget the denominator entirely.
Trap 03
Sign error in the quotient rule numerator
The quotient rule is $u'v - uv'$, not $u'v + uv'$. The minus sign matters and a common error is to use plus instead.

Think & type: Explain in your own words why the derivative of a product $uv$ has two terms, not one.

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Quick-fire practice · 5 problems

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Drill it, then lock it in

Run the quick drill and copy the summary into your book.

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Differentiate $y = xe^x$ using the product rule.

2

Differentiate $y = \dfrac{x^2}{x + 3}$ using the quotient rule.

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Differentiate $y = (x + 1)(x^2 - 2)$. Could you simplify first?

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Differentiate $y = \dfrac{e^x}{x}$.

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Find the gradient of $y = x^2 e^x$ at $x = 0$.

Fill the blanks: drag each token into the matching blank.

u'v uv' v² minus

The product rule gives $y' = $ ___ $+$ ___. The quotient rule numerator uses a ___ sign, and the denominator is ___.

Match each function to its derivative.

  • $y = x^2 e^x$
  • $y = \dfrac{x}{x+1}$
  • $y = (2x+1)e^x$
  • $2e^x + (2x+1)e^x$
  • $\dfrac{1}{(x+1)^2}$
  • $2xe^x + x^2e^x$
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Revisit your thinking

Earlier you were asked what you know about differentiating products and quotients. The key insight is that the derivative of a product is not the product of the derivatives. The product rule adds the two cross-terms because both functions contribute simultaneously to the rate of change. The quotient rule captures the same idea for division, with the extra $v^2$ denominator accounting for the scaling effect.

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Multiple choice

Answer the drill bank and rate your confidence.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

Write full responses, then check them against the model answers.

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Short answer
ApplyBand 4

Q1. Differentiate $y = x^3 e^x$. Show all working. 3 MARKS

View comprehensive answer

Let $u = x^3$ and $v = e^x$ [0.5]. $u' = 3x^2$ and $v' = e^x$ [0.5]. $y' = 3x^2e^x + x^3e^x$ [2].

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ApplyBand 4

Q2. Differentiate $y = \dfrac{2x + 1}{x - 3}$. Show all working. 3 MARKS

View comprehensive answer

Let $u = 2x + 1$ and $v = x - 3$ [0.5]. $u' = 2$ and $v' = 1$ [0.5]. $y' = \frac{2(x-3) - (2x+1)(1)}{(x-3)^2} = \frac{-7}{(x-3)^2}$ [2].

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AnalyseBand 5

Q3. Differentiate $y = x^2 (3x+1)^4$. Identify which rules you used at each step. 4 MARKS

View comprehensive answer

Product rule needed because $x^2$ and $(3x+1)^4$ are multiplied [0.5]. Let $u = x^2$ ($u' = 2x$) and $v = (3x+1)^4$ [0.5]. For $v'$, use chain rule: $v' = 12(3x+1)^3$ [1.5]. $y' = 2x(3x+1)^4 + 12x^2(3x+1)^3$ [1.5].

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Module overview · Maths Advanced · Checkpoint 1