Year 11 Maths Advanced Module 3 ⏱ ~40 min Lesson 5 of 15

The Chain Rule

What do a clock pendulum, a satellite orbiting Earth, and a sound wave all have in common? Their behaviour is described by functions inside functions. The chain rule is the key to differentiating these composite functions — and it is one of the most important tools in all of calculus.

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Think First

Imagine you are driving a car. Your fuel consumption depends on your speed, and your speed depends on how hard you press the accelerator. If you want to know how pressing the accelerator affects fuel use, you need to combine two rates of change: how speed changes with the pedal, and how fuel use changes with speed. How do you think these two rates might be connected?

Type your initial response below — you will revisit this at the end of the lesson.

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📐

Formula Reference — This Lesson

Chain rule (function notation)
$$\frac{d}{dx}\bigl(f(g(x))\bigr) = f'(g(x)) \cdot g'(x)$$
Chain rule (Leibniz notation)
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ where $y = f(u)$ and $u = g(x)$
Procedure
1. Identify the outer function and the inner function 2. Differentiate the outer function (leave the inner alone) 3. Multiply by the derivative of the inner function
Key insight: The chain rule tells us that the rate of change of a composite function is the product of the rates of change of its parts.
📖 Know

Key Facts

  • The chain rule formula in both function and Leibniz notation
  • How to identify the "outer" and "inner" functions
  • That composite functions require the chain rule
💡 Understand

Concepts

  • Why the derivative of a composite function involves multiplication of derivatives
  • How the chain rule extends the power rule to composite functions
  • The relationship between nested processes and their combined rate of change
✅ Can Do

Skills

  • Differentiate composite polynomials using the chain rule
  • Differentiate functions involving brackets raised to a power
  • Apply the chain rule in combination with other differentiation rules
  • Use the chain rule in physical and geometric contexts

Misconceptions to Fix

Wrong: √(a + b) = √a + √b.

Right: The square root of a sum is not the sum of square roots; √(a+b) cannot be simplified this way.

Key Terms
DerivativeThe rate of change of a function at a point; the gradient of the tangent.
DifferentiationThe process of finding the derivative of a function.
Stationary PointA point where the derivative equals zero.
Chain RuleA rule for differentiating composite functions: dy/dx = dy/du × du/dx.
Product RuleA rule for differentiating products: d(uv)/dx = u(dv/dx) + v(du/dx).
Chain RuleFor composite functions: dy/dx = dy/du * du/dx.
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What Is the Chain Rule?

Many functions are not simple polynomials — they are composite, meaning one function is inside another. For example:

$$y = (3x^2 + 1)^5$$

Here, the inner function is $u = 3x^2 + 1$, and the outer function is $y = u^5$.

The chain rule says:

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

In words: differentiate the outer function with respect to the inner function, then multiply by the derivative of the inner function with respect to $x$.

Alternatively, in function notation:

$$\frac{d}{dx}\bigl(f(g(x))\bigr) = f'(g(x)) \cdot g'(x)$$

Why this matters for economics. A company's profit depends on sales, and sales depend on advertising spend. To find how profit changes with advertising, economists use the chain rule: they multiply the rate at which profit changes with sales by the rate at which sales change with advertising. This helps businesses allocate budgets where they have the biggest impact.
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Applying the Chain Rule Step by Step

Here is a reliable procedure for using the chain rule:

  1. Identify the inner function $u = g(x)$ and the outer function $y = f(u)$.
  2. Differentiate the outer function with respect to $u$ to get $\frac{dy}{du}$.
  3. Differentiate the inner function with respect to $x$ to get $\frac{du}{dx}$.
  4. Multiply them together: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
  5. Substitute back any $u$ expressions in terms of $x$.

With practice, you will perform these steps mentally. But at first, write them out clearly — this prevents the most common chain rule errors.

🧮 Worked Examples

Worked Example 1 — Chain Rule with a Polynomial Inner Function

Stepwise
Differentiate $y = (2x + 3)^4$.
  1. 1
    Identify inner and outer functions
    u = 2x + 3 \quad \text{and} \quad y = u^4
  2. 2
    Differentiate outer function with respect to $u$
    \frac{dy}{du} = 4u^3
  3. 3
    Differentiate inner function with respect to $x$
    \frac{du}{dx} = 2
  4. 4
    Apply the chain rule
    \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3 \cdot 2 = 8u^3
  5. 5
    Substitute back $u = 2x + 3$
    \frac{dy}{dx} = 8(2x + 3)^3
✓ Answer $\frac{dy}{dx} = 8(2x + 3)^3$

Worked Example 2 — Chain Rule with a Quadratic Inner Function

Stepwise
Differentiate $y = (x^2 + 5x)^6$.
  1. 1
    Identify inner and outer functions
    u = x^2 + 5x \quad \text{and} \quad y = u^6
  2. 2
    Differentiate each part
    \frac{dy}{du} = 6u^5 \quad \text{and} \quad \frac{du}{dx} = 2x + 5
  3. 3
    Multiply and substitute back
    \frac{dy}{dx} = 6u^5 \cdot (2x + 5) = 6(x^2 + 5x)^5(2x + 5)
✓ Answer $\frac{dy}{dx} = 6(x^2 + 5x)^5(2x + 5)$

Worked Example 3 — Chain Rule Combined with Other Rules

Stepwise
Differentiate $y = 3x^2 + (4x - 1)^5$.
  1. 1
    Differentiate term by term
    The first term $3x^2$ uses the power rule directly. The second term $(4x - 1)^5$ requires the chain rule.
  2. 2
    First term
    \frac{d}{dx}(3x^2) = 6x
  3. 3
    Second term (chain rule)
    u = 4x - 1, \quad y = u^5
    \frac{dy}{du} = 5u^4, \quad \frac{du}{dx} = 4
    \frac{d}{dx}\bigl((4x - 1)^5\bigr) = 5u^4 \cdot 4 = 20(4x - 1)^4
  4. 4
    Combine
    \frac{dy}{dx} = 6x + 20(4x - 1)^4
✓ Answer $\frac{dy}{dx} = 6x + 20(4x - 1)^4$
⚠️

Common Mistakes — Don't Lose Easy Marks

Forgetting to multiply by the derivative of the inner function
Students often stop after differentiating the outer function, giving answers like $\frac{dy}{dx} = 4(2x + 3)^3$ instead of $8(2x + 3)^3$.
✓ Fix: Always ask: "Did I multiply by the derivative of what's inside the brackets?"
Trying to expand brackets before differentiating
Expanding $(3x^2 + 1)^5$ is impossibly long and unnecessary. The chain rule exists precisely to avoid this.
✓ Fix: For composite functions, use the chain rule. Do not expand.
Confusing inner and outer functions
In $y = (x^2 + 1)^3$, the outer function is the cube, not the quadratic.
✓ Fix: Ask yourself: "What is the last operation performed on $x$?" That is the outer function.

📓 Copy Into Your Books

🔗 Chain rule

  • $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
  • $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

📝 Procedure

  • Identify inner $u$ and outer $y$
  • Differentiate outer, keep inner
  • Multiply by derivative of inner

⚠️ Watch out

  • Don't forget $g'(x)$ at the end
  • Don't expand — use the rule

💡 Tip

  • The last operation on $x$ is the outer function

📝 How are you completing this lesson?

🧪 Activities

🔍 Activity 1 — Calculate

Use the Chain Rule

Differentiate each function using the chain rule. Show your working.

  1. 1 $y = (3x + 2)^5$

    Type your answer:

    Answer in your workbook.

    Answer in your workbook
  2. 2 $f(x) = (x^2 - 4)^3$

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    Answer in your workbook.

    Answer in your workbook
  3. 3 $y = (5x^2 + 3x)^4$

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    Answer in your workbook.

    Answer in your workbook
  4. 4 $f(x) = 2x^3 + (x + 1)^6$

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    Answer in your workbook.

    Answer in your workbook
🎨 Activity 2 — Interpret

Explain the Chain Rule

Use words to explain what is happening.

  1. 1 Why do we need the chain rule for $y = (2x + 1)^3$, but not for $y = x^3 + 1$?

    Type your answer:

    Answer in your workbook.

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  2. 2 A student differentiates $y = (x^2 + 3)^4$ and gets $4(x^2 + 3)^3$. What did they forget, and why does the correct answer need that extra factor?

    Type your answer:

    Answer in your workbook.

    Answer in your workbook
Revisit Your Thinking

Earlier you thought about how pressing the accelerator affects fuel consumption.

The connection is exactly the chain rule: the rate of change of fuel with respect to the accelerator equals the rate of change of fuel with respect to speed, multiplied by the rate of change of speed with respect to the accelerator. This is why the chain rule uses multiplication — the effects are chained together, with each rate scaling the next.

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.

Annotate your initial response in your book
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Interactive: Chain Rule Decomposer
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✍️ Short Answer

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Extended Questions

ApplyBand 4

6. (a) Find $\frac{dy}{dx}$ for $y = (2x^2 + 3x)^5$. 2 MARKS

(b) Find the gradient of the tangent to $y = (2x^2 + 3x)^5$ at $x = 1$. 2 MARKS

Type your answers below:

Answer in your workbook.

✏️ Answer in your workbook
AnalyseBand 5

7. A balloon is being inflated so that its radius $r$ (in cm) at time $t$ (in seconds) is given by $r(t) = 3t + 1$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Use the chain rule to find the rate of change of volume with respect to time at $t = 2$ seconds, and explain what this value means. 4 MARKS

Type your answer below:

Answer in your workbook.

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AnalyseBand 5

8. Explain why the derivative of $f(g(x))$ involves multiplication rather than addition of $f'$ and $g'$. Use a real-world example (such as gears, or the fuel example from the lesson) to support your explanation. 3 MARKS

Type your answer below:

Answer in your workbook.

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✅ Comprehensive Answers

🔍 Activity 1 — Calculate Model Answers

1. $\frac{dy}{dx} = 5(3x + 2)^4 \cdot 3 = 15(3x + 2)^4$

2. $f'(x) = 3(x^2 - 4)^2 \cdot 2x = 6x(x^2 - 4)^2$

3. $\frac{dy}{dx} = 4(5x^2 + 3x)^3 \cdot (10x + 3)$

4. $f'(x) = 6x^2 + 6(x + 1)^5$

🎨 Activity 2 — Interpret Model Answers

1. $y = (2x + 1)^3$ is a composite function (a function inside a function), so the chain rule is needed. $y = x^3 + 1$ is just a sum of two simple functions, so we differentiate term by term without the chain rule.

2. The student forgot to multiply by the derivative of the inner function $x^2 + 3$, which is $2x$. The correct answer is $4(x^2 + 3)^3 \cdot 2x = 8x(x^2 + 3)^3$. Without this factor, the rate of change of the inner function is ignored.

❓ Multiple Choice

1. B — Using the chain rule: $\frac{dy}{dx} = 4(3x+2)^3 \\cdot 3 = 12(3x+2)^3$.

2. B — $\frac{dy}{dx} = 3(x^2+1)^2 \\cdot 2x = 6x(x^2+1)^2$.

3. B — The chain rule is $\frac{d}{dx}f(g(x)) = f'(g(x)) \\cdot g'(x)$.

4. B — We write $y = \\sqrt{u}$ where $u = 5x + 1$. The inner function is $5x + 1$.

5. A — Chain rule: $6(x^2-3x)^5 \\cdot (2x - 3) = 6(x^2 - 3x)^5(2x - 3)$.

6. B — $\frac{dy}{dx} = 3(4x-1)^2 \\cdot 4 = 12(4x-1)^2$.

7. B — $\frac{dy}{dx} = 2(x^3+2) \\cdot 3x^2 = 6x^2(x^3+2)$.

📝 Short Answer Model Answers

Q6 (4 marks): (a) Let $u = 2x^2 + 3x$, so $y = u^5$. Then $\frac{dy}{du} = 5u^4$ and $\frac{du}{dx} = 4x + 3$ [1]. By the chain rule, $\frac{dy}{dx} = 5(2x^2 + 3x)^4(4x + 3)$ [1]. (b) At $x = 1$, $\frac{dy}{dx} = 5(2 + 3)^4(4 + 3) = 5 \times 625 \times 7 = 21875$ [2].

Q7 (4 marks): $\frac{dV}{dr} = 4\pi r^2$ and $\frac{dr}{dt} = 3$ [1]. By the chain rule, $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 12\pi r^2$ [1]. At $t = 2$, $r = 3(2) + 1 = 7$, so $\frac{dV}{dt} = 12\pi(49) = 588\pi$ cm$^3$/s [1]. This means that after 2 seconds, the balloon's volume is increasing at a rate of $588\pi$ cubic centimetres per second [1].

Q8 (3 marks): The chain rule uses multiplication because the rates of change are compounded through each stage [1]. For example, fuel consumption depends on speed, and speed depends on accelerator position. A small change in the accelerator produces a change in speed, which then produces a change in fuel use [1]. The total effect is the product of the individual rates of change, not their sum [1].

Consolidation Game

The Chain Rule

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