Lesson 3: Product & Quotient Rules

Lesson at a Glance | He Tirohanga Whakamua

Do NowWhy the derivative of a product is NOT simply the product of derivatives10 min

Product Rule MasteryApplying

\frac{d}{d x} [u v] = u^{'} v + u v^{'}

to algebraic functions15 min

Quotient Rule MasteryApplying

\frac{d}{d x} [\frac{u}{v}] = \frac{u^{'} v - u v^{'}}{v^{2}}

and simplifying numerators15 min

Portfolio EntryWrite Section 3: Product & Quotient Rule Problem Solver10 min

Exit DrillDifferentiate

y = \frac{3 x^{2} + 1}{2 x - 5}

5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

The Product Rule: $\frac{d}{d x} [u (x) \cdot v (x)] = u^{'} (x) v (x) + u (x) v^{'} (x)$ .
The Quotient Rule: $\frac{d}{d x} [\frac{u (x)}{v (x)}] = \frac{u^{'} (x) v (x) - u (x) v^{'} (x)}{[v (x)]^{2}}$ .
Algebraic simplification techniques for factoring out common terms from product/quotient derivative expressions.

Students will demonstrate

By executing product and quotient derivatives combined with chain rule terms.
By completing Section 3 of their Level 3 Calculus Differentiation Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Product Trap Prompt:

"Consider $y = x^{2} \cdot x^{3} = x^{5}$ . Using the power rule directly, $\frac{d y}{d x} = 5 x^{4}$ . What happens if you incorrectly take derivative of $x^{2}$ ( $2 x$ ) times derivative of $x^{3}$ ( $3 x^{2}$ )?"

Unpack: You get $2 x \cdot 3 x^{2} = 6 x^{3}$ , which is WRONG! Product derivatives must follow $(u v)^{'} = u^{'} v + u v^{'}$ . Here: $(2 x) (x^{3}) + (x^{2}) (3 x^{2}) = 2 x^{4} + 3 x^{4} = 5 x^{4}$ ! The product rule works!

Product & Quotient Rule Worked Examples (15 min)

1. Product Rule Example

$y = (3 x - 1)^{4} (2 x + 5)^{3}$
$u^{'} = 12 (3 x - 1)^{3}, v^{'} = 6 (2 x + 5)^{2}$
$\frac{d y}{d x} = 12 (3 x - 1)^{3} (2 x + 5)^{3} + 6 (3 x - 1)^{4} (2 x + 5)^{2}$ .

2. Quotient Rule Example

$y = \frac{x^{2} + 3}{4 x - 1}$
$\frac{d y}{d x} = \frac{(2 x) (4 x - 1) - (x^{2} + 3) (4)}{(4 x - 1)^{2}} = \frac{8 x^{2} - 2 x - 4 x^{2} - 12}{(4 x - 1)^{2}} = \frac{4 x^{2} - 2 x - 12}{(4 x - 1)^{2}}$ .

📁 Calculus Differentiation Portfolio — Section 3: Product & Quotient Rules

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

Section 3 Requirements:

1. Product vs Quotient Comparison Box: Write down both formulas clearly, highlighting the minus sign in quotient numerator and squared denominator.

2. Product/Quotient Solver Set: Differentiate $y = (x^{3} - 2 x) \sqrt{4 x + 1}$ and $y = \frac{(2 x + 1)^{3}}{x^{2} - 4}$ , simplifying numerators fully.

3. Alternative Method Rationale: 1-paragraph explanation comparing quotient rule vs rewriting $\frac{u}{v} = u \cdot v^{- 1}$ using product + chain rule.

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 = (6x² − 30x − 2)/(2x − 5)², with x ≠ 5/2.
Portfolio 1: (uv)′ = u′v + uv′ and (u/v)′ = (u′v − uv′)/v², where v ≠ 0.
Portfolio 2: The derivatives are (14x³ + 3x² − 12x − 2)/√(4x + 1), for x > −1/4, and 2(2x + 1)²(x − 4)(x + 3)/(x² − 4)², for x ≠ ±2.
Portfolio 3: Differentiating uv⁻¹ gives u′v⁻¹ − uv′v⁻² = (u′v − uv′)/v²; both forms require v ≠ 0.

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 3 differentiates y = (3x^2+1)/(2x-5) yielding dy/dx = (6x^2 - 30x - 2)/(2x-5)^2 via quotient rule."

Teacher Planning & NCEA Alignment

NCEA Level 3 Calculus Alignment (6 Credits External):

Differentiation Methods: Apply the Product Rule $(u v)^{'} = u^{'} v + u v^{'}$ and Quotient Rule $(\frac{u}{v})^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ to algebraic functions.

Vocabulary: Product rule, quotient rule, derivative of product, numerator simplification, common factor.