Mathematics Advanced • Year 11 • Module 3 • Lesson 5

The Chain Rule

HSC-style writing on the chain rule — short answers plus an extended response combining decomposition, tangent equations and gradient analysis.

Master · Past-Paper Style

1. Short-answer questions

1.1 Differentiate y = (5x − 2)⁷. Show the u-substitution and chain-rule step. 2 marks Band 3

1.2 Differentiate y = cos(3x² + 1). Show the u-substitution and chain-rule step, and present the final answer in terms of x. 3 marks Band 3-4

1.3 Find the gradient of the curve y = (x³ − 2)² at the point where x = 1. 3 marks Band 4

Stuck on 1.3? Find dy/dx via chain rule first, then substitute x = 1.

2. Extended response

2.1 Consider the curve y = (x² − 4x + 5)³.
(a) Using the chain rule (showing the decomposition u = x² − 4x + 5 and y = u³), find dy/dx in terms of x.
(b) Find the gradient of the tangent to the curve at the point where x = 2.
(c) Find the y-coordinate at x = 2, and hence the equation of the tangent at that point.
(d) Find the x-coordinates of all points on the curve at which the tangent is horizontal.
(e) Explain in 1-2 sentences why the curve has only one horizontal tangent, even though dy/dx contains the factor 3 u², which could in principle vanish for several values of x. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — u = x² − 4x + 5, dy/du = 3u², du/dx = 2x − 4.

• 1 mark — dy/dx = 3 u² (2x − 4) = 3 (2x − 4) (x² − 4x + 5)², with u substituted back.

Part (b) — 1 mark

• 1 mark — at x = 2: dy/dx = 3 · (0) · (1)² = 0.

Part (c) — 1 mark

• 1 mark — y(2) = (4 − 8 + 5)³ = 1³ = 1; point (2, 1); tangent y − 1 = 0 (x − 2) ⇒ y = 1.

Part (d) — 2 marks

• 1 mark — sets dy/dx = 3 (2x − 4) (x² − 4x + 5)² = 0 and analyses both factors.

• 1 mark — shows x² − 4x + 5 = (x − 2)² + 1 > 0 for all x, so (x² − 4x + 5)² ≠ 0; hence only 2x − 4 = 0 ⇒ x = 2.

Part (e) — 1 mark

• 1 mark — explains: the inner quadratic x² − 4x + 5 has positive discriminant test (Δ = 16 − 20 = −4 < 0), so it has no real zeros and never equals zero; only the linear factor 2x − 4 can vanish, at the single value x = 2.

Your response:

Stuck on (e)? Compute the discriminant of x² − 4x + 5 — what does it tell you about real roots?

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — y = (5x − 2)⁷ (2 marks)

Sample response. Let u = 5x − 2, so y = u⁷. dy/du = 7 u⁶; du/dx = 5. dy/dx = 7 u⁶ · 5 = 35 u⁶ = 35 (5x − 2)⁶.

Marking notes. 1 mark — identifies u and computes dy/du and du/dx correctly. 1 mark — multiplies and substitutes u back. Bare answer "35(5x − 2)⁶" with no working scores 1/2. Common error: writing 7(5x − 2)⁶ without the × 5 (Trap 01 in the lesson).

1.2 — y = cos(3x² + 1) (3 marks)

Sample response. Let u = 3x² + 1, so y = cos u. dy/du = −sin u; du/dx = 6x. dy/dx = (−sin u) · (6x) = −6x sin u = −6x sin(3x² + 1).

Marking notes. 1 mark — correct decomposition. 1 mark — correct derivatives of outer (−sin u) and inner (6x), with the negative sign carried. 1 mark — final answer in terms of x (not u). Common errors: missing the negative sign from differentiating cos; leaving the answer in terms of u (Trap 03 in the lesson) — loses 1 mark.

1.3 — Gradient of y = (x³ − 2)² at x = 1 (3 marks)

Sample response. Let u = x³ − 2, so y = u². dy/du = 2u; du/dx = 3x². dy/dx = 2u · 3x² = 6x² (x³ − 2). At x = 1: 6 · 1 · (1 − 2) = 6 · (−1) = −6.

Marking notes. 1 mark — correct dy/du and du/dx. 1 mark — correct dy/dx in terms of x. 1 mark — correct numerical evaluation at x = 1 (with sign). Common error: writing (1 − 2) = 1 (sign flip) — loses 0.5 to 1 mark.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a). Let u = x² − 4x + 5, so y = u³. dy/du = 3 u²; du/dx = 2x − 4. By the chain rule, dy/dx = (3 u²)(2x − 4) = 3 (2x − 4) (x² − 4x + 5)². [1 mark — decomposition; 1 mark — chain rule applied with u substituted back.]

Part (b). At x = 2: 2x − 4 = 0, so dy/dx = 3 · 0 · (x² − 4x + 5)² = 0. [1 mark.]

Part (c). y(2) = (4 − 8 + 5)³ = (1)³ = 1. Point: (2, 1). Tangent: y − 1 = 0(x − 2) ⇒ y = 1 (horizontal). [1 mark.]

Part (d). Horizontal tangent ⇔ dy/dx = 0 ⇔ 3 (2x − 4) (x² − 4x + 5)² = 0. The constant 3 ≠ 0, so we need either 2x − 4 = 0 or (x² − 4x + 5)² = 0. [1 mark — sets up the factored equation.] The inner quadratic x² − 4x + 5 = (x − 2)² + 1 ≥ 1 > 0 for all real x, so (x² − 4x + 5)² > 0 always — this factor never vanishes. Hence only 2x − 4 = 0 contributes, giving x = 2. [1 mark — eliminates the squared factor and identifies the single x-value.]

Part (e). The inner quadratic x² − 4x + 5 has discriminant Δ = 16 − 20 = −4 < 0, so it has no real zeros and never equals zero. Consequently the squared factor in dy/dx is strictly positive, leaving only the linear factor 2x − 4 to vanish — which it does at exactly one x-value (x = 2). Hence there is only one horizontal tangent. [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Identifies u and dy/du but does not multiply by du/dx (Trap 01); for (b)-(c) computes gradient and tangent for a specific x correctly but stops there. ≈ 2-3 marks.

Band 4: (a)-(c) correct. (d) sets dy/dx = 0 and finds x = 2 from the linear factor but also incorrectly lists x-values from setting the squared quadratic factor to zero (without checking if it actually has real roots). ≈ 4-5 marks.

Band 5: All parts attempted. (d) correctly handles only x = 2 but does not justify why; (e) gives a vague "the squared factor cannot be zero" without invoking the discriminant. ≈ 5-6 marks.

Band 6: Every step shown, "u" substituted back in final answer, completing-the-square or discriminant argument used in (d) and (e) to prove the inner quadratic is always positive. 7/7.