Module 3 · L5 of 15~40 min⚡ +50 XP in Learn · +25 to complete
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.
Today's hook, A car's speed changes with time, and the fuel burn rate depends on speed. If you want to know how fast your fuel is burning right now, you need both rates. That's the chain rule in action. How do you think mathematicians combine these two layers of change?
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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.
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, $(x^2 + 1)^3$ versus $x^2 + 1$ cubed on its own. What do you think would change in the differentiation process?
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Two moves · decompose then multiply
Work through the core explanation before applying it.
$\dfrac{dy}{du}$ = derivative of outer with respect to $u$
$\dfrac{du}{dx}$ = derivative of inner with respect to $x$
In function notation: if $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \cdot g'(x)$.
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What you'll 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'll master
Know
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: $h'(x) = f'(g(x)) \cdot g'(x)$
Understand
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 $x$
Can do
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
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Key terms
Composite function
A function of the form $f(g(x))$ where one function is applied to the result of another.
Chain rule
If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
Inner function
The function $g(x)$ that is substituted into the outer function $f$.
Outer function
The function $f$ that acts on the output of the inner function.
$u$-substitution
Naming the inner function $u = g(x)$ to make chain rule steps explicit and trackable.
Back-substitution
Replacing $u$ with $g(x)$ after differentiating, so the final answer is in terms of $x$ only.
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The chain rule, the idea
Work through the core explanation before applying it.
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The chain rule, the idea
core concept · +3 XP at end
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.
Think of it like a gear system. The overall gear ratio is the product of the ratios of each pair of connected gears. If the first gear multiplies speed by 3, and the second multiplies by 2, the total multiplication is $3 \times 2 = 6$. The chain rule works the same way: multiply the derivative 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 $y = (2x^3 + 5)^4$, which is the correct inner function $u$?
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Power rule chain, the shortcut
core concept
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:
If $y = [g(x)]^n$, then $\dfrac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$. This is the power rule combined with the chain rule. You bring down the power, reduce it by one, and multiply by the derivative of whatever is inside the brackets.
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
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Work examples end to end
Follow the reasoning through complete worked solutions.
Worked example 1 · polynomial composite+5 XP on full reveal
Differentiate $y = (3x + 2)^5$.
Band 3Apply
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Let $u = 3x + 2$, so $y = u^5$
Identify the inner function $u = 3x + 2$ and the outer function $y = u^5$.
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$\dfrac{dy}{du} = 5u^4$ and $\dfrac{du}{dx} = 3$
Differentiate the outer function with respect to $u$, and the inner function with respect to $x$.
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$\dfrac{dy}{dx} = 5u^4 \cdot 3 = 15(3x + 2)^4$
Apply the chain rule: multiply the derivatives, then back-substitute $u = 3x + 2$.
Worked example 2 · power composite+5 XP on full reveal
Differentiate $y = (2x^3 + 5)^4$.
Band 4Apply
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Let $u = 2x^3 + 5$, so $y = u^4$
The inner function is $u = 2x^3 + 5$ and the outer function is $y = u^4$.
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$\dfrac{dy}{du} = 4u^3$ and $\dfrac{du}{dx} = 6x^2$
Differentiate: derivative of $u^4$ is $4u^3$; derivative of $2x^3 + 5$ is $6x^2$.
Substitute $x = 1$ to find the gradient at that specific point.
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.
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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.
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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.
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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.
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Drill it, then lock it in
Run the quick drill and copy the summary into your book.
Odd one out: Which of these does NOT require the chain rule?
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Build the idea
Work through the core explanation before applying it.
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Revisit your thinking
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 $x$.
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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.