Not every function is a simple power or a composition inside brackets. Some functions are products of two simpler functions. Others are quotients — one function divided by another. To differentiate these, we need two more powerful tools: the product rule and the quotient rule.
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Consider $f(x) = x^2 \cdot x^3$. You already know this simplifies to $x^5$, so its derivative is $5x^4$. But what about $f(x) = x^2 \cdot (3x + 1)$? You could expand it, but what if the functions were more complicated — like $x^2 \cdot \sqrt{x+1}$ — where expansion is impossible? How do you think we might differentiate a product of two functions without expanding?
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Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: The derivative of a quotient is the quotient of the derivatives.
Right: The quotient rule is (u/v)' = (u'v - uv')/v². The derivative of a quotient is NOT simply u'/v'.
📚 Core Content
If $y = u \cdot v$, where $u$ and $v$ are both functions of $x$, then:
$$\frac{dy}{dx} = u'v + uv'$$
Intuitively, this makes sense: the rate of change of a product depends on how fast the first part changes (while the second stays the same) plus how fast the second part changes (while the first stays the same).
Differentiate $y = x^2(3x + 1)$.
Let $u = x^2$ and $v = 3x + 1$. Then $u' = 2x$ and $v' = 3$.
$$\frac{dy}{dx} = (2x)(3x + 1) + (x^2)(3) = 6x^2 + 2x + 3x^2 = 9x^2 + 2x$$
We can check this by expanding first: $y = 3x^3 + x^2$, so $\frac{dy}{dx} = 9x^2 + 2x$ ✓.
If $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$, then:
$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$$
Many students remember this as: "Bottom times derivative of top, minus top times derivative of bottom, over bottom squared."
Differentiate $y = \frac{2x + 1}{x - 3}$.
Let $u = 2x + 1$ and $v = x - 3$. Then $u' = 2$ and $v' = 1$.
$$\frac{dy}{dx} = \frac{2(x - 3) - (2x + 1)(1)}{(x - 3)^2} = \frac{2x - 6 - 2x - 1}{(x - 3)^2} = \frac{-7}{(x - 3)^2}$$
Sometimes the product or quotient rule must be used together with the chain rule. For example, if $y = x^2(3x + 1)^4$, the product rule gives:
$$\frac{dy}{dx} = (2x)(3x + 1)^4 + x^2 \cdot \frac{d}{dx}\bigl((3x + 1)^4\bigr)$$
The second term requires the chain rule: $\frac{d}{dx}\bigl((3x + 1)^4\bigr) = 4(3x + 1)^3 \cdot 3 = 12(3x + 1)^3$.
So the complete derivative is:
$$\frac{dy}{dx} = 2x(3x + 1)^4 + 12x^2(3x + 1)^3$$
🧮 Worked Examples
🧪 Activities
1 $y = (x + 3)(2x - 1)$
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2 $f(x) = \frac{3x}{x + 2}$
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3 $y = x^2(4x - 1)^3$
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4 $f(x) = \frac{x^2 + 1}{2x - 3}$
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1 A student claims: "Because $(x \cdot x)' = (x^2)' = 2x$, and $x' \cdot x' = 1 \cdot 1 = 1$, the product rule must be wrong." Explain the flaw in this argument.
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2 Explain why the quotient rule has $v^2$ in the denominator rather than just $v$.
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Earlier you thought about differentiating a product without expanding.
The product rule does exactly this: it tells us that the derivative of $u \cdot v$ is $u'v + uv'$. This accounts for how both factors change simultaneously. For something like $x^2 \cdot \sqrt{x+1}$, expansion is impossible — but the product rule combined with the chain rule gives us the answer directly. That is why these rules are essential.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Find the derivative of each of the following. Show all working. 4 MARKS
(a) $y = (2x + 1)(x^2 - 3x)$
(b) $y = \frac{4x - 1}{x + 2}$
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7. A curve has equation $y = \frac{x^2}{x + 1}$. Find the coordinates of the point(s) where the tangent to the curve is horizontal. 4 MARKS
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8. A company's revenue is modelled by $R(x) = x(50 - x)$, where $x$ is the price per unit. Find the rate of change of revenue with respect to price when $x = 20$, and interpret what this means for the business. 3 MARKS
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1. $y' = (1)(2x - 1) + (x + 3)(2) = 2x - 1 + 2x + 6 = 4x + 5$
2. $f'(x) = \frac{3(x + 2) - 3x(1)}{(x + 2)^2} = \frac{6}{(x + 2)^2}$
3. $y' = (2x)(4x - 1)^3 + x^2 \cdot 3(4x - 1)^2 \cdot 4 = 2x(4x - 1)^3 + 12x^2(4x - 1)^2$
4. $f'(x) = \frac{2x(2x - 3) - (x^2 + 1)(2)}{(2x - 3)^2} = \frac{4x^2 - 6x - 2x^2 - 2}{(2x - 3)^2} = \frac{2x^2 - 6x - 2}{(2x - 3)^2}$
1. The coincidence that $(x^2)' = 2x$ does not prove a general rule. The product rule gives $(x \cdot x)' = (1)(x) + (x)(1) = 2x$, which agrees. But for general functions like $x \cdot \sin x$, multiplying derivatives gives the wrong answer.
2. The $v^2$ arises because differentiating $\frac{u}{v}$ is equivalent to differentiating $u \cdot v^{-1}$, and applying the product rule produces $v^2$ in the denominator after combining terms over a common denominator.
1. B — Product rule: $(1)(3x^2) + (x)(6x) = 3x^2 + 6x^2 = 9x^2$.
2. A — Quotient rule: $\frac{2(x+1) - (2x-1)(1)}{(x+1)^2} = \frac{3}{(x+1)^2}$.
3. B — The product rule is $(uv)' = u'v + uv'$.
4. A — Quotient rule: $\frac{1(x-2) - (x+1)(1)}{(x-2)^2} = \frac{-3}{(x-2)^2}$.
5. A — Product rule: $(2x)(x+3) + (x^2)(1) = 2x^2 + 6x + x^2 = 3x^2 + 6x$.
6. A — Quotient rule: $\frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$.
7. A — Product rule plus chain rule: $(1)(2x-1)^3 + x \\cdot 3(2x-1)^2 \\cdot 2 = (2x-1)^3 + 6x(2x-1)^2 = 2x(2x-1)^3 + 6x^2(2x-1)^2$ after factoring or expanding.
Q6 (4 marks): (a) Using the product rule with $u = 2x + 1$ and $v = x^2 - 3x$: $u' = 2$, $v' = 2x - 3$ [1]. $\frac{dy}{dx} = 2(x^2 - 3x) + (2x + 1)(2x - 3) = 6x^2 - 10x - 3$ [1]. (b) Using the quotient rule with $u = 4x - 1$ and $v = x + 2$: $u' = 4$, $v' = 1$ [1]. $\frac{dy}{dx} = \frac{4(x + 2) - (4x - 1)(1)}{(x + 2)^2} = \frac{9}{(x + 2)^2}$ [1].
Q7 (4 marks): Using the quotient rule: $\frac{dy}{dx} = \frac{2x(x + 1) - x^2(1)}{(x + 1)^2} = \frac{x^2 + 2x}{(x + 1)^2}$ [2]. For a horizontal tangent, $\frac{dy}{dx} = 0$, so $x^2 + 2x = 0$, giving $x = 0$ or $x = -2$ [1]. When $x = 0$, $y = 0$; when $x = -2$, $y = -4$. The points are $(0, 0)$ and $(-2, -4)$ [1].
Q8 (3 marks): $R(x) = 50x - x^2$, so $\frac{dR}{dx} = 50 - 2x$ [1]. At $x = 20$, $\frac{dR}{dx} = 50 - 40 = 10$ [1]. This means that when the price is \$20, a \$1 increase in price will increase revenue by approximately \$10, so raising the price is currently beneficial [1].
The Product and Quotient Rules
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