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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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.
Write your initial response in your book. You will revisit it at the end of the lesson.
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.
📚 Core Content
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:
$$\frac{d}{dx}\bigl(f(g(x))\bigr) = f'(g(x)) \cdot g'(x)$$
Here is a reliable procedure for using the chain rule:
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
🧪 Activities
1 $y = (3x + 2)^5$
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2 $f(x) = (x^2 - 4)^3$
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3 $y = (5x^2 + 3x)^4$
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4 $f(x) = 2x^3 + (x + 1)^6$
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1 Why do we need the chain rule for $y = (2x + 1)^3$, but not for $y = x^3 + 1$?
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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?
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Answer in your workbook.
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.
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. (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
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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
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Answer in your workbook.
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
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Answer in your workbook.
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. $f'(x) = 6x^2 + 6(x + 1)^5$
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.
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 — $\frac{dy}{dx} = 2(x^3+2) \\cdot 3x^2 = 6x^2(x^3+2)$.
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].
The Chain Rule
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