Mathematics Advanced • Year 11 • Module 3 • Lesson 4

Differentiation Rules

HSC-style writing on the power, constant-multiple and sum/difference rules — short answers plus an extended response with a proof of the power rule.

Master · Past-Paper Style

1. Short-answer questions

1.1 Differentiate f(x) = x⁴ − 3x² + 2/x − √x . Show all working, including the rewrite of each non-polynomial term as a power. 4 marks Band 4

1.2 Find the gradient of the tangent to y = x³ − 2x² + 4 at the point where x = −1. 3 marks Band 3-4

1.3 A curve has equation y = x³ − 6x² + 9x + 1. Find the x-coordinates of all points on the curve where the tangent has zero gradient. 3 marks Band 4

Stuck on 1.3? dy/dx = 0 leads to a quadratic — factor it.

2. Extended response

2.1 Consider the function f(x) = 2x³ − 9x² + 12x + 5.
(a) Find f ′(x).
(b) Find all x-values at which the tangent to y = f(x) has zero gradient (stationary points).
(c) Find the corresponding y-values.
(d) Use first principles to prove the power rule d/dx (x³) = 3x² for the specific case n = 3, using the expansion (x + h)³ = x³ + 3x²h + 3xh² + h³.
(e) Briefly explain (in 1-2 sentences) why this single-case derivation is enough to illustrate the general power rule, and why the binomial theorem is what extends it to any positive integer n. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 1 mark

• 1 mark — f ′(x) = 6x² − 18x + 12 (every term correct).

Part (b) — 2 marks

• 1 mark — sets f ′(x) = 0 and factors 6x² − 18x + 12 = 6(x − 1)(x − 2).

• 1 mark — solutions x = 1 and x = 2.

Part (c) — 1 mark

• 1 mark — y(1) = 10 and y(2) = 9 both computed correctly; stationary points stated as (1, 10) and (2, 9).

Part (d) — 2 marks

• 1 mark — sets up the first principles definition with f(x) = x³, expands (x + h)³ as given, simplifies the numerator to 3x²h + 3xh² + h³.

• 1 mark — divides by h, simplifies to 3x² + 3xh + h², takes lim_h→0 to obtain 3x².

Part (e) — 1 mark

• 1 mark — explains that the same pattern (one leading nxⁿ⁻¹h term, all higher-order h terms vanishing in the limit) extends to general positive integer n via the binomial expansion (x + h)ⁿ = xⁿ + n xⁿ⁻¹ h + (higher h-terms).

Your response:

Stuck on (d)? Follow the four-step process: f(x + h), simplify numerator, factor h, take limit.

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What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Differentiate f(x) = x⁴ − 3x² + 2/x − √x (4 marks)

Sample response. Rewrite in index form: f(x) = x⁴ − 3x² + 2 x⁻¹ − x^1/2. Differentiate term by term using the power rule:

d/dx (x⁴) = 4x³; d/dx (−3x²) = −6x; d/dx (2 x⁻¹) = −2 x⁻²; d/dx (−x^1/2) = −(1/2) x^−1/2.

Combine: f ′(x) = 4x³ − 6x − 2/x² − 1 / (2 √x).

Marking notes. 1 mark — correct rewrite of all non-polynomial terms in index form. 1 mark — correct derivative of polynomial terms (4x³, −6x). 1 mark — correct derivative of −1 and 1/2 power terms (with signs). 1 mark — final answer in either index form or radical/reciprocal form. Common errors: forgetting the negative sign on −2/x² (sign comes from −1 in exponent); cancelling the 1/2 against the x^−1/2.

1.2 — Tangent gradient to y = x³ − 2x² + 4 at x = −1 (3 marks)

Sample response. dy/dx = 3x² − 4x. At x = −1: dy/dx = 3(1) − 4(−1) = 3 + 4 = 7.

Marking notes. 1 mark — correct derivative. 1 mark — correct substitution, especially the (−1)² = 1 and −4(−1) = +4 sign handling. 1 mark — final value 7 stated. Common error: writing 3(−1)² = −3 (treating (−1)² as −1) — this loses 1 mark.

1.3 — Stationary points of y = x³ − 6x² + 9x + 1 (3 marks)

Sample response. dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Set = 0: x = 1 or x = 3.

Marking notes. 1 mark — correct derivative. 1 mark — correct factoring (most efficient: factor out 3 first, then factorise the trinomial). 1 mark — both solutions stated. Acceptable alternatives: using the quadratic formula on 3x² − 12x + 9 = 0 to obtain the same roots.

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

Sample Band-6 response.

Part (a). Differentiate term by term: f ′(x) = 2 · 3x² − 9 · 2x + 12 = 6x² − 18x + 12. [1 mark.]

Part (b). Stationary ⇔ f ′(x) = 0 ⇔ 6x² − 18x + 12 = 0 ⇔ 6(x² − 3x + 2) = 0 ⇔ 6(x − 1)(x − 2) = 0. [1 mark — factoring.] Hence x = 1 or x = 2. [1 mark — both solutions.]

Part (c). y(1) = 2 − 9 + 12 + 5 = 10; y(2) = 16 − 36 + 24 + 5 = 9. Stationary points: (1, 10) and (2, 9). [1 mark.]

Part (d). Let f(x) = x³. Then f(x + h) = (x + h)³ = x³ + 3x²h + 3xh² + h³ (given).

f(x + h) − f(x) = 3x²h + 3xh² + h³ = h(3x² + 3xh + h²).

[1 mark — expansion and numerator simplification.]

( f(x + h) − f(x) ) / h = 3x² + 3xh + h² (h ≠ 0).

f ′(x) = lim_h→0 (3x² + 3xh + h²) = 3x² + 0 + 0 = 3x². [1 mark — cancel, limit, conclusion.]

Part (e). The same pattern works for any positive integer n: by the binomial theorem, (x + h)ⁿ = xⁿ + n xⁿ⁻¹ h + (terms with h², h³, ..., hⁿ). Subtracting xⁿ and dividing by h gives n xⁿ⁻¹ + (terms with at least one h still remaining); every term except the first vanishes in the limit h → 0, leaving f ′(x) = n xⁿ⁻¹ — the general power rule. [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Differentiates (a), attempts (b) but factoring is incomplete (uses quadratic formula but stops mid-calculation); does not attempt (d) or only writes the first principles formula without execution. ≈ 3-4 marks.

Band 4: Parts (a)-(c) correct. Sets up (d) and reaches 3x² + 3xh + h² but does not take the limit; (e) attempts but vague. ≈ 4-5 marks.

Band 5: All numerical parts clean. (d) complete including the limit; (e) names the binomial theorem but does not explain why the higher-order h-terms vanish. ≈ 5-6 marks.

Band 6: Every step clean; (d) carries "h ≠ 0" annotation when cancelling; (e) explicitly states "every higher-order h-term contains a positive power of h and therefore vanishes as h → 0". 7/7.