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📚Unit Plans → 📐Level 3 Calculus: Differentiation → 1First Principles
Year 13 (NCEA Level 3) Pāngarau / Mathematics

Lesson 1: Derivative Foundations & First Principles

NCEA Level 3 Calculus. Students master secant-to-tangent limits, Newton's first-principles derivative definition f′(x)=limh→0f(x+h)−f(x)h, and power rules, writing Portfolio Section 1.

Lesson at a Glance | He Tirohanga Whakamua

Do NowCalculating average velocity vs finding exact instantaneous velocity at t=3s10 min
First Principles Limit DefinitionExpanding f′(x)=limh→0f(x+h)−f(x)h algebraically15 min
Power Rule & Polynomial DerivativesApplying ddx[xn]=nxn−1 for negative and fractional exponents15 min
Portfolio EntryWrite Section 1: First Principles & Limit Derivative Map10 min
Exit DrillDifferentiate f(x)=3x2−5x+4 from first principles5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

  • The difference between average rate of change (secant gradient ΔyΔx) and instantaneous rate of change (tangent gradient f′(x)).
  • The Newton-Leibniz First Principles Definition:
    f′(x)=limh→0f(x+h)−f(x)h
  • The Power Rule: ddx[axn]=a·nxn−1 for all real powers n∈ℝ.

Students will demonstrate

  • By deriving f′(x)=2x for f(x)=x2 algebraically without skipping limit cancellation steps.
  • By completing Section 1 of their Level 3 Calculus Differentiation Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Instantaneous Rate of Change Prompt:

"If a car travels 100 km in 2 hours, its average speed is 50 km/h. But a speed camera catches the driver doing 95 km/h at exactly t=42min. How do we calculate the exact gradient of a curve at a single instantaneous point?"

Unpack: By bringing two points on the curve infinitely close together! We take a secant line between (x,f(x)) and (x+h,f(x+h)), and evaluate the limit as the distance h→0.

First Principles Algebraic Proof (15 min)

1. First Principles Step-by-Step
  1. Write f(x+h)−f(x)h
  2. Expand (x+h)2=x2+2xh+h2
  3. Cancel x2: 2xh+h2h
  4. Factor h: h(2x+h)h=2x+h
  5. Take limit h→0→f′(x)=2x.
2. General Power Rule (ddx[xn]=nxn−1)

• f(x)=x5→f′(x)=5x4
• f(x)=1x2=x−2→f′(x)=−2x−3=−2x3
• f(x)=x=x1/2→f′(x)=12x−1/2=12x.

📁 Calculus Differentiation Portfolio — Section 1: First Principles

Students open their Level 3 Calculus Portfolio and complete Section 1:

Section 1 Requirements:

1. Secant to Tangent Geometry Diagram: Sketch a curve y=f(x), drawing secants with decreasing interval h, showing how the limit yields the tangent slope f′(x).

2. First Principles Derivation: Fully derive f′(x) for f(x)=2x2−3x+1 showing all limit expansion steps for h→0.

3. Power Rule Drill: Differentiate y=4x3−3x+5x3 expressing answers with positive rational exponents.

Kaiako answer and checking guide

Scope: Mathematical checking support for the shipped tasks. This is not an NZQA marking schedule or an assessment-evidence requirement.

  1. Exit drill: For f(x) = 3x2 − 5x + 4, f(x + h) − f(x) = h(6x + 3h − 5). After dividing by h and taking h → 0, f′(x) = 6x − 5.
  2. Portfolio 1: The diagram should identify P = (x, f(x)), Q = (x + h, f(x + h)), the secant slope [f(x + h) − f(x)]/h, and the tangent obtained as Q approaches P.
  3. Portfolio 2: The difference quotient for 2x2 − 3x + 1 simplifies to 4x + 2h − 3, so the limit is 4x − 3.
  4. Portfolio 3: dy/dx = 12x2 + 3/x2 + (15/2)x1/2. For the real-valued expression as written, x > 0.

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 1 derives f'(x) = 4x - 3 from first principles and differentiates negative/fractional powers using the power rule."

Teacher Planning & NCEA Alignment

NCEA Level 3 Calculus Alignment (6 Credits External):

  • Differentiation Methods: Demonstrate understanding of rate of change, limits from first principles, and power rule polynomial differentiation.

Vocabulary: Derivative, first principles, limit (limh→0), secant line, tangent line, instantaneous rate of change, power rule.