Mathematics Advanced • Year 11 • Module 3 • Lesson 5

The Chain Rule

Build procedural fluency with the chain rule: identify the inner function u, the outer function f(u), differentiate each, then multiply.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the chain rule in Leibniz notation:

If y = f(u) and u = g(x), then dy/dx = ____________ · ____________ .

Q1.2 For each composite function, identify the inner function u and the outer function f(u):

(a) y = (3x + 2)⁵ → u = ________ ; f(u) = ________

(b) y = sin(2x) → u = ________ ; f(u) = ________

Q1.3 True or false: applying the chain rule to y = x⁵ (no inner function) gives a wrong answer. T / F Why: ____________________________________________

Stuck? Revisit lesson § Key Terms and § Two moves.

2. Worked example — differentiate y = (3x + 2)⁵

Follow each line. Every step has a reason on the right.

Problem. Differentiate y = (3x + 2)⁵.

Step 1 — Decompose.

Let u = 3x + 2. Then y = u⁵.

Reason: the (3x + 2) is the inner function; the 5th power is applied outside.

Step 2 — Differentiate each layer.

dy/du = 5u⁴ (power rule on outer)

du/dx = 3 (power rule on inner; constant disappears)

Step 3 — Apply the chain rule (multiply).

dy/dx = dy/du · du/dx = 5u⁴ · 3 = 15 u⁴

Step 4 — Substitute u back in terms of x.

dy/dx = 15 (3x + 2)⁴

Reason: HSC requires the final answer in terms of x, not u.

Conclusion. dy/dx = 15 (3x + 2)⁴.

3. Faded example — fill in the missing steps

Differentiate y = (x² + 1)⁴. Fill in each blank. 3 marks

Step 1 — Decompose:

Let u = ________________ , so y = u^____ .

Step 2 — Differentiate each:

dy/du = ____ · u^____ du/dx = ________

Step 3 — Multiply:

dy/dx = ____ · u^____ · ____ = ________ · u^____

Step 4 — Substitute back:

dy/dx = ________ ( ________________ )^____

Stuck? Revisit lesson § Worked Example 3.

4. Graduated practice — differentiate using the chain rule

State u and dy/du, du/dx, then write dy/dx in terms of x.

Foundation — linear inner function (4 questions)

Q	Function	dy/dx
4.1 1	y = (2x + 1)³
4.2 1	y = (5x − 2)⁷
4.3 1	y = cos(4x)
4.4 1	y = sin(3x − 1)

Standard — typical HSC difficulty (6 questions)

Show u, du/dx, dy/du, then multiply.

4.5 y = (x² − 3)⁶. 2 marks

4.6 y = sin(x² + 1). 2 marks

4.7 y = sin(2x³). 2 marks

4.8 y = (x³ − 2x)⁴. 2 marks

4.9 Find the gradient of y = (3x − 1)⁴ at x = 1. 2 marks

4.10 Find the gradient of y = cos(3x²) at x = 0. 2 marks

Extension — combine concepts (2 questions)

4.11 Differentiate y = √(x² + 4) by first rewriting as (x² + 4)^1/2. 3 marks

4.12 Find the equation of the tangent to y = (x² + 1)³ at the point where x = 1. 3 marks

Stuck on 4.12? Find dy/dx at x = 1 (gradient), then the point (1, 8), then point-gradient form.

5. Self-check the easy 3

Tick the first three once you have checked your method works.

For 4.1 I multiplied by du/dx = 2 (do not forget that × 2!).

For 4.3 I differentiated cos(u) as −sin(u), then multiplied by du/dx = 4 (and kept the negative sign).

I always substituted u back to x in my final answer (no "u" left).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Chain rule

dy/dx = dy/du · du/dx.

Q1.2 — Inner / outer identification

(a) u = 3x + 2, f(u) = u⁵. (b) u = 2x, f(u) = sin u. (c) u = x² − 1, f(u) = u⁴.

Q1.3 — Chain rule on y = x⁵

F (false). If we "set u = x", then u = g(x) = x and du/dx = 1, so dy/dx = 5u⁴ · 1 = 5x⁴ — the correct answer, just with an unnecessary extra layer. The chain rule does not give a wrong answer; it is simply not needed when there is no genuine inner function (Trap 02 in the lesson).

Q3 — Faded example y = (x² + 1)⁴

Step 1: u = x² + 1, y = u⁴. Step 2: dy/du = 4 · u³; du/dx = 2x. Step 3: dy/dx = 4 · u³ · 2x = 8x · u³. Step 4: dy/dx = 8x (x² + 1)³.

Q4.1 — y = (2x + 1)³

u = 2x + 1; dy/du = 3u²; du/dx = 2. dy/dx = 3u² · 2 = 6 (2x + 1)².

Q4.2 — y = (5x − 2)⁷

u = 5x − 2; dy/du = 7u⁶; du/dx = 5. dy/dx = 35 (5x − 2)⁶.

Q4.3 — y = cos(4x)

u = 4x; dy/du = −sin u; du/dx = 4. dy/dx = −4 sin(4x).

Q4.4 — y = sin(3x − 1)

u = 3x − 1; dy/du = cos u; du/dx = 3. dy/dx = 3 cos(3x − 1).

Q4.5 — y = (x² − 3)⁶

u = x² − 3; dy/du = 6 u⁵; du/dx = 2x. dy/dx = 6 u⁵ · 2x = 12x (x² − 3)⁵.

Q4.6 — y = sin(x² + 1)

u = x² + 1; dy/du = cos u; du/dx = 2x. dy/dx = 2x cos(x² + 1).

Q4.7 — y = sin(2x³)

u = 2x³; dy/du = cos u; du/dx = 6x². dy/dx = 6x² cos(2x³).

Q4.8 — y = (x³ − 2x)⁴

u = x³ − 2x; dy/du = 4 u³; du/dx = 3x² − 2. dy/dx = 4 (3x² − 2)(x³ − 2x)³.

Q4.9 — Gradient of y = (3x − 1)⁴ at x = 1

dy/dx = 4(3x − 1)³ · 3 = 12 (3x − 1)³. At x = 1: 12 · (2)³ = 12 · 8 = 96.

Q4.10 — Gradient of y = cos(3x²) at x = 0

dy/dx = −sin(3x²) · 6x = −6x sin(3x²). At x = 0: −0 · sin(0) = 0. (Horizontal tangent at x = 0.)

Q4.11 — y = √(x² + 4)

Rewrite: y = (x² + 4)^1/2. u = x² + 4; dy/du = (1/2) u^−1/2; du/dx = 2x. dy/dx = (1/2) u^−1/2 · 2x = x · u^−1/2 = x / √(x² + 4).

Q4.12 — Tangent to y = (x² + 1)³ at x = 1

dy/dx = 3(x² + 1)² · 2x = 6x (x² + 1)². At x = 1: gradient = 6 · 1 · (2)² = 24. Point: y = (1 + 1)³ = 8, so (1, 8). Tangent: y − 8 = 24(x − 1) ⇒ y = 24x − 16.