The Chain Rule
Differentiating composite functions: when one function is inside another, the rate of change depends on both layers. A thermostat adjusts heating based on temperature, which itself changes with time — the rate of heating depends on the rate of temperature change.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Before we start, what do you already know about functions inside functions? Think about what it means for one function to be "inside" another — for example, $\sin(x^2)$ versus $\sin(x)$ squared. What do you think would change in the differentiation process?
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$
In function notation: if $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \cdot g'(x)$.
Key facts
- the chain rule formula: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
- how to identify inner and outer functions
- the function notation form: $h'(x) = f'(g(x)) \cdot g'(x)$
Concepts
- why differentiation of composite functions requires two layers
- how the $u$-substitution makes the chain rule systematic
- why the answer must always be back-substituted into $x$
Skills
- apply the chain rule to polynomial and power composites
- apply the chain rule to trigonometric composites
- find the gradient of a curve at a point using the chain rule
If $y$ depends on $u$, and $u$ depends on $x$, then $y$ depends on $x$ through two layers. The chain rule tells us that the rate of change of $y$ with respect to $x$ is the product of the rates through each layer.
In function notation: if $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \cdot g'(x)$.
Notice that $f'$ is evaluated at $g(x)$ — the outer function's derivative is computed at the inner function's value, not at $x$ directly. This is what makes the chain rule different from simple power differentiation.
The two-step method
- Decompose: name the inner function $u = g(x)$ and express $y = f(u)$.
- Differentiate each layer: find $\frac{dy}{du}$ and $\frac{du}{dx}$, then multiply.
- Back-substitute: replace $u$ with $g(x)$ to get the answer in terms of $x$.
Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ — multiply rates through each layer; Step 1: let $u$ = inner function; write $y$ as a function of $u$
Pause — copy the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ and the setup procedure (let $u$ = inner function, write $y$ as a function of $u$) into your book.
Did you get this? True or false: to apply the chain rule to $y = (3x + 2)^5$, you let $u = 3x + 2$ so the outer function becomes $y = u^5$.
Quick check: For the function $y = \sin(2x^3)$, which is the correct inner function $u$?
We just saw that the chain rule means multiplying $\dfrac{dy}{du}$ by $\dfrac{du}{dx}$ — differentiating each layer and multiplying. That raises a question: when the outer function is a power, can we skip the substitution step and write the answer directly? This card answers it → the power chain shortcut: $[g(x)]^n \to n[g(x)]^{n-1} \cdot g'(x)$.
When the outer function is a power and the inner function is a polynomial, the chain rule produces a recognisable pattern:
For example, $y = (3x + 2)^5$:
- Bring down the power: $5(3x + 2)^4$
- Multiply by the derivative of the inner function: $\times 3$
- Result: $\dfrac{dy}{dx} = 15(3x + 2)^4$
With practice, you can apply this shortcut in one step. Until then, always write out $u$, $\frac{dy}{du}$, and $\frac{du}{dx}$ explicitly.
Power chain shortcut: if $y = [g(x)]^n$, then $y' = n[g(x)]^{n-1} \cdot g'(x)$; Three steps: bring down power → reduce power by 1 → multiply by derivative of inside
Pause — copy the power chain shortcut formula $y' = n[g(x)]^{n-1} \cdot g'(x)$ and the three-step mnemonic (bring down, reduce, multiply by inside derivative) into your book.
Quick check: Using the power chain shortcut, what is $\dfrac{dy}{dx}$ for $y = (2x + 1)^3$?
Worked examples · reveal as you go
Differentiate $y = (3x + 2)^5$.
Differentiate $y = \sin(2x^3)$.
Find the gradient of $y = (x^2 + 1)^4$ at $x = 1$.
Common errors · the 3 traps that cost marks
Forgetting to multiply by the derivative of the inner function
Students often write $\frac{dy}{dx} = 5(3x+2)^4$ for $y = (3x+2)^5$, forgetting to multiply by $\frac{du}{dx} = 3$. The chain rule requires multiplying by $\frac{du}{dx}$ every single time — without exception.
✓ Fix: Always write out $\frac{du}{dx}$ before combining. The extra factor is easy to drop when rushing.
Applying the chain rule when it is not needed
For $y = x^5$, there is no inner function — just a straight power. Applying the chain rule here wastes time and can introduce errors. Check whether the argument of the outer function contains more than just $x$ before choosing a rule.
✓ Fix: Ask "is there anything other than a plain $x$ inside?" If no, use the power rule directly.
Leaving the answer in terms of $u$
After applying the chain rule, always substitute $u = g(x)$ back into the final answer. Answers written in terms of $u$ are not accepted in HSC exams — the marker cannot award the method mark unless the answer is in terms of $x$.
✓ Fix: Circle the back-substitution step in your working as a reminder to never skip it.
Activity 1 — Quick-fire practice · 5 problems
Apply the chain rule to each function. Show $u$, $\frac{dy}{du}$, and $\frac{du}{dx}$ in your working.
Differentiate $y = (2x + 1)^3$.
Differentiate $y = \cos(4x)$.
Differentiate $y = (x^2 - 3)^6$.
Find the gradient of $y = (3x - 1)^4$ at $x = 1$.
Differentiate $y = \sin(x^2 + 1)$.
Odd one out: Which of these does NOT require the chain rule?
Earlier you were asked: What do you already know about functions inside functions?
The chain rule shows that differentiating a composite function requires differentiating the outer function while keeping the inner function intact, then multiplying by the derivative of the inner function. This is because change propagates through each layer: the $u$-substitution makes both layers explicit, and back-substitution ensures the final answer is in terms of $x$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Chain rule with polynomial. Differentiate $y = (5x - 2)^7$. (2 marks)
View comprehensive answer
Let $u = 5x - 2$, so $y = u^7$ [0.5]
$\dfrac{dy}{du} = 7u^6$ and $\dfrac{du}{dx} = 5$ [0.5]
$\dfrac{dy}{dx} = 7u^6 \cdot 5 = 35(5x - 2)^6$ [1]
Q2. Chain rule with trigonometric function. Differentiate $y = \cos(3x^2 + 1)$. (3 marks)
View comprehensive answer
Let $u = 3x^2 + 1$, so $y = \cos u$ [0.5]
$\dfrac{dy}{du} = -\sin u$ and $\dfrac{du}{dx} = 6x$ [1]
$\dfrac{dy}{dx} = -\sin u \cdot 6x = -6x\sin(3x^2 + 1)$ [1.5]
Q3. Gradient at a point. Find the gradient of the curve $y = (x^3 - 2)^2$ at the point where $x = 1$. (3 marks)
View comprehensive answer
Let $u = x^3 - 2$, so $y = u^2$ [0.5]
$\dfrac{dy}{du} = 2u$ and $\dfrac{du}{dx} = 3x^2$ [0.5]
$\dfrac{dy}{dx} = 2u \cdot 3x^2 = 6x^2(x^3 - 2)$ [1]
At $x = 1$: $\dfrac{dy}{dx} = 6(1)^2(1 - 2) = -6$ [1]
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