Mathematical Methods · Units 3 & 4

The Chain Rule

Master the chain rule the easy way, with plain English intuition, the inside and outside trick, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Russian dolls are the secret to this whole topic. A function tucked neatly inside another function is called a composite function, and the chain rule is how you differentiate it without unpacking the dolls one painful layer at a time. Once you can spot the function hiding inside, differentiating something like $(3x+1)^5$ takes a few seconds instead of expanding a fifth power by hand. This is one of the most heavily tested ideas in the whole course, so it pays to nail it.

What a composite function really is

Some functions are built by stacking one operation on top of another. Take $(3x + 1)^5$ . To work it out for a number you first do $3x + 1$ , and then you raise that result to the power of five. Two steps, one inside the other. The inside function is the part you do first, here $3x + 1$ . The outside function is what you do to that result, here the fifth power.

Spotting these two layers is ninety per cent of the battle. Ask yourself the simple question, what would I type first on a calculator. That part is the inside.

The chain rule formula

Here is the whole idea in one line. We give the inside function a name, $u$ , so we can talk about the two layers separately.

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

Read it as a chain of rates. The rate of $y$ against $x$ equals the rate of $y$ against the inside, multiplied by the rate of the inside against $x$ . The clever part is that the $du$ symbols look like they cancel, which is a handy way to remember the formula.

The composite y = (3x+1)^5: the inside 3x+1 fixes where the curve crosses zero, at x = -1/3, while the outside fifth power makes it flat near there and very steep away from it.

How to actually do it

The idea is the hard part. The method is a short recipe you repeat every time.

Name the inside function $u$ , and write $y$ in terms of $u$ .
Find $\dfrac{dy}{du}$ by differentiating the outside layer.
Find $\dfrac{du}{dx}$ by differentiating the inside layer.
Multiply the two results, then substitute the inside back in for $u$ .

For a power of a bracket the shortcut is quick to see. To differentiate $(\text{something})^n$ you bring the power down, drop the power by one, and then multiply by the derivative of the something:

\frac{d}{dx}\big[g(x)\big]^n = n\big[g(x)\big]^{n-1} \cdot g'(x)

The trap that catches the most students in real exams is leaving off that final factor $g'(x)$ . Examiner reports note again and again that students omit the inside derivative, the part that sits in front of the bracket. The inside derivative is not optional. A second common slip is dropping the power by the wrong amount, so always reduce the power by exactly one.

Learn it, then fade the help

1 · I do — fully worked

Differentiate $y = (3x + 1)^5$ . Let $u = 3x + 1$ , so $y = u^5$ . Then $\dfrac{dy}{du} = 5u^4$ and $\dfrac{du}{dx} = 3$ . Multiply and put the inside back: $\dfrac{dy}{dx} = 5u^4 \cdot 3 = 15(3x+1)^4$ .

2 · We do — your turn, guided

Same steps, your turn. Differentiate $y = (5x - 2)^4$ . Let $u = 5x - 2$ , so $\dfrac{dy}{du} = 4u^3$ and $\dfrac{du}{dx} = 5$ . Multiply: $\dfrac{dy}{dx} = 4(5x-2)^3 \cdot \;?\;$ .

Check your step

$\dfrac{dy}{dx} = 4(5x-2)^3 \cdot 5 = 20(5x-2)^3$ — the missing factor is the inside derivative $5$ .

3 · You do — solo

No scaffolding now: differentiate $y = (x^2 - 4x)^3$ .

Reveal solution

Let $u = x^2 - 4x$ , so $\dfrac{dy}{du} = 3u^2$ and $\dfrac{du}{dx} = 2x - 4$ . Then $\dfrac{dy}{dx} = 3(x^2 - 4x)^2 (2x - 4)$ — keep that inner factor $(2x-4)$ .

Roots are just powers in disguise

A root is nothing new. Before you differentiate, rewrite every root as a power. A square root is a power of one half, so $\sqrt{2x+7}$ becomes $(2x+7)^{1/2}$ , and a fraction like $\dfrac{1}{(3x+2)^2}$ becomes $(3x+2)^{-2}$ . Once the root or fraction is written as a power, the same four step recipe handles it. Watch the negative signs carefully when the power is negative, because that is where careless errors creep in.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Write the chain rule in Leibniz form.

How do you differentiate a power of a function,

y = (f(x))^n

When do you reach for the chain rule?

What is the single most common chain rule mistake?

How do you differentiate a square root like

\sqrt{2x+7}

Recall · Product and Quotient Rules

Differentiate a product

y = u(x)\,v(x)

Recall · Derivatives of Exponential and Logarithmic Functions

Differentiate

y = \log_e(f(x))

See this recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1A natural logarithm, from a real exam

Let $f(x) = \log_e\left(x^3 - 3x + 2\right)$ . Find $f'(3)$ .

1
The outside is the natural log and the inside is the cubic. Differentiating $\log_e(u)$ gives $\tfrac{1}{u}$ , then multiply by the derivative of the inside.
$f'(x) = \frac{1}{x^3 - 3x + 2} \cdot (3x^2 - 3) = \frac{3x^2 - 3}{x^3 - 3x + 2}$
2
Substitute $x = 3$ into the derivative.
$f'(3) = \frac{3(9) - 3}{27 - 9 + 2} = \frac{24}{20}$
3
Simplify. The examiner report flags omitting the numerator, the inside derivative, as the common error.
$f'(3) = \frac{6}{5}$

Answer

f'(3) = \dfrac{6}{5}

VCAA 2024 Mathematical Methods Exam 1, Q1b

Worked Example 2A root, then a gradient, from a real exam

Let $f(x) = 6\sqrt{x + 1} + 5$ . Find the gradient of the tangent to $y = f(x)$ at $x = 8$ .

1
Write the root as a power first. The constant $+5$ differentiates to zero.
$f(x) = 6(x + 1)^{1/2} + 5$
2
Apply the chain rule to the power. The inside is $x + 1$ , whose derivative is $1$ .
$f'(x) = 6 \cdot \tfrac{1}{2}(x + 1)^{-1/2} = \frac{3}{\sqrt{x + 1}}$
3
The gradient of the tangent at $x = 8$ is the value of $f'(8)$ .
$f'(8) = \frac{3}{\sqrt{9}} = \frac{3}{3} = 1$

Answer

f'(8) = 1

VCAA 2025 Mathematical Methods Exam 1, Q1b

Worked Example 3A power of a linear expression

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

1
Spot the inside and the outside. The inside is $u = 3x + 1$ , the outside is the fifth power.
$y = u^5, \quad u = 3x + 1$
2
Differentiate each piece on its own.
$\frac{dy}{du} = 5u^4, \quad \frac{du}{dx} = 3$
3
Multiply them together, then put the inside back in.
$\frac{dy}{dx} = 5u^4 \cdot 3 = 15(3x+1)^4$

Answer

\frac{dy}{dx} = 15(3x+1)^4

Worked Example 4A power of a polynomial

Differentiate $y = (x^2 - 4x)^3$ .

1
The inside is $u = x^2 - 4x$ , the outside is the cube.
$y = u^3, \quad u = x^2 - 4x$
2
Differentiate the outside and the inside separately.
$\frac{dy}{du} = 3u^2, \quad \frac{du}{dx} = 2x - 4$
3
Multiply, then substitute $u = x^2 - 4x$ back in. Keep the inner factor.
$\frac{dy}{dx} = 3(x^2 - 4x)^2 (2x - 4)$

Answer

\frac{dy}{dx} = 3(x^2 - 4x)^2 (2x - 4)

Worked Example 5A square root

Differentiate $y = \sqrt{2x + 7}$ .

1
A root is a power. Rewrite the root as a power of one half first.
$y = (2x + 7)^{1/2}$
2
Let $u = 2x + 7$ , so the outside is the power $\tfrac{1}{2}$ .
$\frac{dy}{du} = \tfrac{1}{2} u^{-1/2}, \quad \frac{du}{dx} = 2$
3
Multiply and tidy. The $\tfrac{1}{2}$ and the $2$ cancel.
$\frac{dy}{dx} = \tfrac{1}{2}(2x+7)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x+7}}$

Answer

\frac{dy}{dx} = \dfrac{1}{\sqrt{2x+7}}

Practice test

Try it yourself

9 questions, 12 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A differentiable function $f$ satisfies $f(4) = 25$ and $f'(4) = 15$ . The gradient of the tangent to $y = \sqrt{f(x)}$ at $x = 4$ is:

1mark

Need a hint?

Write

y = (f(x))^{1/2}

and use the chain rule. You will need both

f(4)

and

f'(4)

VCAA 2024 Mathematical Methods Exam 2, Section A Q16

Q2.Consider $h(x) = a\log_e(bx)$ , where $a, b \in R \setminus \{0\}$ . Given that the derivative $h'(x)$ has range $(0, \infty)$ , which of the following must be true?

1mark

Need a hint?

Differentiate with the chain rule and watch what happens to

b

VCAA 2025 Mathematical Methods Exam 2, Section A Q16

Q3.Let $f(x) = \sin(x)\,e^{2x}$ . Find $f'\left(\dfrac{\pi}{4}\right)$ .

2marks

Work this on paper. The worked solution appears once you submit.

VCAA 2023 Mathematical Methods Exam 1, Q1b

Q4.The derivative of $y = (5x - 2)^4$ is:

1mark

Need a hint?

Let

u = 5x - 2

. Differentiate the outside

u^4

and the inside

5x - 2

separately, then multiply them.

Q5.The derivative of $y = (x^2 + 3)^6$ is:

1mark

Q6.If $f(x) = \sqrt{4x - 3}$ , then $f'(x)$ is:

1mark

Q7.For $y = (2x - 1)^3$ , the value of $\dfrac{dy}{dx}$ when $x = 1$ is:

1mark

Q8.The derivative of $y = \dfrac{1}{(3x+2)^2}$ is:

1mark

Q9.Let $f(x) = (x^2 - 6x)^4$ . Find $f'(x)$ and the value of $f'(3)$ . Show every step.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

When do I use the chain rule?

Whenever you differentiate a function tucked inside another function, a composite function, such as a bracket raised to a power, a root, or e to the power of something other than just x.

What is the most common chain rule mistake?

Forgetting to multiply by the derivative of the inside. After differentiating the outside, you must multiply by the derivative of whatever was inside.

How do I differentiate a square root using the chain rule?

Rewrite the root as a power of one half, then apply the chain rule. For example, the square root of (2x + 7) becomes (2x + 7) to the power one half, and differentiating gives one over the square root of (2x + 7).

The Chain Rule

Learn

What a composite function really is

The chain rule formula

How to actually do it

Roots are just powers in disguise

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions