Lesson 2: Chain Rule | Level 3 Calculus

Lesson at a Glance | He Tirohanga Whakamua

Do NowWhy expanding

(3 x^{2} + 5)^{8}

by hand is insane and why we need a chain rule10 min

Function DecompositionIdentifying inner

u = g (x)

and outer

y = f (u)

components15 min

Chain Rule MasteryApplying

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

and power shortcut

n u^{n - 1} u^{'}

15 min

Portfolio EntryWrite Section 2: Chain Rule & Inner-Outer Function Decomposition10 min

Exit DrillDifferentiate

y = (4 x^{3} - 7 x)^{6}

5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

How to decompose a composite function $y = f (g (x))$ into inner variable $u = g (x)$ and outer function $y = f (u)$ .
The Chain Rule Formula:
$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ or $\frac{d}{d x} [(g (x))^{n}] = n (g (x))^{n - 1} \cdot g^{'} (x)$ .
How to differentiate fractional powers of composite functions such as $y = \sqrt{5 x^{2} - 3 x + 1} = (5 x^{2} - 3 x + 1)^{1 / 2}$ .

Students will demonstrate

By executing 5 multi-step chain rule derivatives including negative/fractional powers.
By completing Section 2 of their Level 3 Calculus Differentiation Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Composite Power Prompt:

"To differentiate $y = (2 x + 1)^{2}$ , you can expand to $y = 4 x^{2} + 4 x + 1$ and get $\frac{d y}{d x} = 8 x + 4$ . But how would you differentiate $y = (2 x + 1)^{100}$ without expanding 101 terms?"

Unpack: By using the Chain Rule! Let $u = 2 x + 1 \to y = u^{100}$ . Then $\frac{d y}{d u} = 100 u^{99}$ and $\frac{d u}{d x} = 2$ . Multiplying yields $\frac{d y}{d x} = 100 (2 x + 1)^{99} \cdot 2 = 200 (2 x + 1)^{99}$ !

Chain Rule Worked Examples (15 min)

1. Polynomial Power Example

$y = (4 x^{3} - 7 x)^{6}$
$u = 4 x^{3} - 7 x \to \frac{d u}{d x} = 12 x^{2} - 7$
$\frac{d y}{d x} = 6 (4 x^{3} - 7 x)^{5} \cdot (12 x^{2} - 7)$ .

2. Radical Expression Example

$y = \sqrt{3 x^{2} + 4} = (3 x^{2} + 4)^{1 / 2}$
$\frac{d y}{d x} = \frac{1}{2} (3 x^{2} + 4)^{- 1 / 2} \cdot (6 x) = \frac{3 x}{\sqrt{3 x^{2} + 4}}$ .

📁 Calculus Differentiation Portfolio — Section 2: Chain Rule

Students open their Level 3 Calculus Portfolio and complete Section 2:

Section 2 Requirements:

1. Function Decomposition Matrix: Complete a table listing 4 composite functions, identifying inner $u = g (x)$ , outer $y = f (u)$ , $\frac{d u}{d x}$ , and $\frac{d y}{d u}$ .

2. Chain Rule Problem Set: Differentiate $y = (2 x^{4} - 5 x^{2} + 3)^{8}$ and $y = \frac{1}{(4 x - 1)^{3}}$ .

3. Extension: Fractional Radical Rationale: 1-paragraph explanation of how rewriting roots as fractional exponents enables the chain rule to differentiate complex engineering stress functions.

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.

Exit drill: dy/dx = 6(4x³ − 7x)⁵(12x² − 7).
Portfolio 1: Any four valid composite functions can be used. Each row should correctly identify u = g(x), y = f(u), du/dx, dy/du, and their product.
Portfolio 2: The derivatives are 8(2x⁴ − 5x² + 3)⁷(8x³ − 10x) and −12/(4x − 1)⁴.
Portfolio 3: A sound explanation rewrites a root as u^p/q, differentiates the outer power, multiplies by u′, and retains any domain restriction from the original root or denominator.

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 2 differentiates y = (4x^3 - 7x)^6 yielding dy/dx = 6(4x^3 - 7x)^5 (12x^2 - 7) via inner/outer chain rule decomposition."

Teacher Planning & NCEA Alignment

NCEA Level 3 Calculus Alignment (6 Credits External):

Differentiation Methods: Apply the Chain Rule $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ to differentiate composite algebraic and radical functions.

Vocabulary: Chain rule, composite function, inner function ( $u$ ), outer function ( $f (u)$ ), Leibniz notation ( $\frac{d y}{d x}$ ), fractional exponent.

Lesson 2: Chain Rule & Composite Functions