Lesson Note on Differentiation: Rules, Product and Quotient Rule for SSS 2

Class: Senior Secondary School 2 (SS2 / SSS2)
Term: 2nd Term
Week: 3
Age: 16 years
Duration: 45 minutes
Subject: Further Mathematics
Curriculum Theme: Differentiation
Previous Lesson: Differentiation: Transcendental Functions.
Topic: DIFFERENTIATION
Subject Matter: Rules of differentiation, product rule, quotient rule, function of function (chain rule)

Specific Objectives

By the end of the lesson, pupils should be able to:

Cognitive Domain:

• Define differentiation as a concept in calculus.
• State the basic rules of differentiation (constant, power, sum/difference, constant multiple).
• State the product rule, quotient rule, and chain rule for differentiation.

Affective Domain:

• Appreciate the importance of differentiation in solving mathematical problems.
• Develop interest in applying differentiation rules to various functions.

Psychomotor Domain:

• Apply the basic rules to differentiate algebraic functions.
• Solve problems involving the product rule accurately.
• Solve problems involving the quotient rule accurately.
• Solve problems involving the chain rule (function of a function) accurately.

Social Domain:

• Collaborate with peers to solve differentiation problems.

Reference Materials

The following resources were used in planning this lesson:

• Senior Secondary Schools Education Curriculum
• State Unified Scheme of Work
• Further Mathematics for Senior Secondary Schools, Book 2.
• https://www.khanacademy.org/math/calculus-1/cs1-differentiation-rules
• https://www.youtube.com/watch?v=k_A6pM-aV0Q

Instructional Materials

The teacher will teach this lesson with the aid of:

• Charts showing the rules of differentiation.
• Whiteboard and markers.
• Further Mathematics textbooks.

Rationale for the Lesson

This lesson helps pupils understand how to find the rate of change of functions, which is important for solving problems in physics, engineering, and economics. It builds a foundation for more advanced calculus topics and practical applications.

Prerequisite/Previous Knowledge

Pupils are expected to have basic knowledge of algebraic expressions, indices, and functions.

Lesson Content/Board Summary

DIFFERENTIATION

Introduction to Differentiation

Differentiation is a process in calculus used to find the derivative or rate of change of a function. The derivative of a function y = f(x) with respect to x is denoted as dy/dx or f'(x).

Basic Rules of Differentiation

• Constant Rule: If y = c (where c is a constant), then dy/dx = 0.
• Power Rule: If y = ax^n, then dy/dx = anx^(n-1).
• Sum/Difference Rule: If y = f(x) ± g(x), then dy/dx = d/dx [f(x)] ± d/dx [g(x)].
• Constant Multiple Rule: If y = c * f(x), then dy/dx = c * d/dx [f(x)].

Example: Differentiate y = 5x^3 – 4x^2 + 7x – 9.

Solution:

Step 1: Differentiate each term separately using the power rule and constant rule.

d/dx (5x^3) = 5 * 3x^(3-1) = 15x^2

d/dx (-4x^2) = -4 * 2x^(2-1) = -8x

d/dx (7x) = 7 * 1x^(1-1) = 7x^0 = 7 * 1 = 7

d/dx (-9) = 0

Step 2: Combine the derivatives.

dy/dx = 15x^2 – 8x + 7

Product Rule

The product rule is used to differentiate a function that is the product of two functions, say y = uv, where u and v are functions of x.

Formula: dy/dx = u(dv/dx) + v(du/dx)

Example 1: Differentiate y = (x^2 + 1)(3x – 2).

Solution:

Step 1: Identify u and v.

Let u = x^2 + 1, and v = 3x – 2.

Step 2: Find du/dx and dv/dx.

du/dx = d/dx (x^2 + 1) = 2x

dv/dx = d/dx (3x – 2) = 3

Step 3: Apply the product rule formula.

dy/dx = u(dv/dx) + v(du/dx)

dy/dx = (x^2 + 1)(3) + (3x – 2)(2x)

Step 4: Simplify the expression.

dy/dx = 3x^2 + 3 + 6x^2 – 4x

dy/dx = 9x^2 – 4x + 3

Example 2: Differentiate y = (2x + 3)(x^3 – 1).

Solution:

Step 1: Identify u and v.

Let u = 2x + 3, and v = x^3 – 1.

Step 2: Find du/dx and dv/dx.

du/dx = d/dx (2x + 3) = 2

dv/dx = d/dx (x^3 – 1) = 3x^2

Step 3: Apply the product rule formula.

dy/dx = u(dv/dx) + v(du/dx)

dy/dx = (2x + 3)(3x^2) + (x^3 – 1)(2)

Step 4: Simplify the expression.

dy/dx = 6x^3 + 9x^2 + 2x^3 – 2

dy/dx = 8x^3 + 9x^2 – 2

Quotient Rule

The quotient rule is used to differentiate a function that is the ratio of two functions, say y = u/v, where u and v are functions of x.

Formula: dy/dx = [v(du/dx) – u(dv/dx)] / v^2

Example 1: Differentiate y = (4x – 1) / (x + 2).

Solution:

Step 1: Identify u and v.

Let u = 4x – 1, and v = x + 2.

Step 2: Find du/dx and dv/dx.

du/dx = d/dx (4x – 1) = 4

dv/dx = d/dx (x + 2) = 1

Step 3: Apply the quotient rule formula.

dy/dx = [(x + 2)(4) – (4x – 1)(1)] / (x + 2)^2

Step 4: Simplify the expression.

dy/dx = [4x + 8 – 4x + 1] / (x + 2)^2

dy/dx = 9 / (x + 2)^2

Example 2: Differentiate y = (x^2 + 3) / (2x – 1).

Solution:

Step 1: Identify u and v.

Let u = x^2 + 3, and v = 2x – 1.

Step 2: Find du/dx and dv/dx.

du/dx = d/dx (x^2 + 3) = 2x

dv/dx = d/dx (2x – 1) = 2

Step 3: Apply the quotient rule formula.

dy/dx = [(2x – 1)(2x) – (x^2 + 3)(2)] / (2x – 1)^2

Step 4: Simplify the expression.

dy/dx = [4x^2 – 2x – 2x^2 – 6] / (2x – 1)^2

dy/dx = (2x^2 – 2x – 6) / (2x – 1)^2

Chain Rule (Function of a Function)

The chain rule is used to differentiate composite functions, i.e., functions within functions. If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). Alternatively, if y = f(u) and u = g(x), then dy/dx = (dy/du) * (du/dx).

Example 1: Differentiate y = (3x + 5)^4.

Solution:

Step 1: Let u be the inner function.

Let u = 3x + 5.

Step 2: Express y in terms of u and find dy/du.

Then y = u^4. So, dy/du = 4u^3.

Step 3: Find du/dx.

du/dx = d/dx (3x + 5) = 3.

Step 4: Apply the chain rule formula dy/dx = (dy/du) * (du/dx).

dy/dx = (4u^3) * (3)

dy/dx = 12u^3

Step 5: Substitute u back into the expression.

dy/dx = 12(3x + 5)^3

Example 2: Differentiate y = sqrt(x^2 – 4).

Solution:

Step 1: Rewrite the function and identify the inner function.

y = (x^2 – 4)^(1/2). Let u = x^2 – 4.

Step 2: Express y in terms of u and find dy/du.

Then y = u^(1/2). So, dy/du = (1/2)u^(-1/2) = 1 / (2 * sqrt(u)).

Step 3: Find du/dx.

du/dx = d/dx (x^2 – 4) = 2x.

Step 4: Apply the chain rule formula dy/dx = (dy/du) * (du/dx).

dy/dx = [1 / (2 * sqrt(u))] * (2x)

dy/dx = 2x / (2 * sqrt(u))

dy/dx = x / sqrt(u)

Step 5: Substitute u back into the expression.

dy/dx = x / sqrt(x^2 – 4)

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 5 minutes
Teaching Skill: Set Induction
Teacher’s Activity: The teacher introduces the topic of differentiation by asking pupils about rates of change in everyday life, e.g., speed as a rate of change of distance. The teacher then defines differentiation.
Pupils’ Activity: Pupils respond to the questions and listen attentively to the introduction.
Learning Point: Pupils recall prior knowledge related to rates of change and are introduced to differentiation.

Step 2: Basic Rules of Differentiation

Time: 10 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains the basic rules of differentiation (constant, power, sum/difference, constant multiple rules) and demonstrates their application with examples on the whiteboard.
Pupils’ Activity: Pupils listen, take notes, and ask questions for clarification.
Learning Point: Pupils understand and can apply the basic rules of differentiation.

Step 3: Product Rule

Time: 8 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher introduces the product rule, writes the formula on the board, and solves two worked examples step-by-step, explaining each stage.
Pupils’ Activity: Pupils copy the formula and examples into their notes and pay attention to the steps.
Learning Point: Pupils learn how to apply the product rule for differentiating products of functions.

Step 4: Quotient Rule

Time: 8 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher introduces the quotient rule, writes the formula, and solves two worked examples step-by-step, emphasizing the order of operations.
Pupils’ Activity: Pupils observe the demonstration, copy the examples, and ask questions.
Learning Point: Pupils learn how to apply the quotient rule for differentiating quotients of functions.

Step 5: Chain Rule (Function of a Function)

Time: 8 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains the concept of a composite function and introduces the chain rule, providing the formula and solving two worked examples.
Pupils’ Activity: Pupils follow the explanation and examples, noting the substitution steps.
Learning Point: Pupils learn how to apply the chain rule for differentiating composite functions.

Step 6: Class Activity/Practice

Time: 5 minutes
Teaching Skill: Guided Practice
Teacher’s Activity: The teacher gives a few practice questions covering all the rules taught and guides pupils as they attempt to solve them.
Pupils’ Activity: Pupils solve the practice questions individually or in pairs.
Learning Point: Pupils reinforce their understanding and application of the differentiation rules.

Step 7: Evaluation/Review

Time: 5 minutes

Teaching Skill: Questioning/Assessment

Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

1. Define differentiation.
2. State the power rule of differentiation.
3. Differentiate y = (5x – 3)(x^2 + 4) using the product rule.
4. Differentiate y = (x^3 + 2) / (x – 1) using the quotient rule.
5. Differentiate y = (2x^2 + 7)^3 using the chain rule.

Pupils’ Activity: Pupils answer orally and in writing.

Learning Point: Pupils demonstrate understanding of the lesson.

Step 8: Conclusion

Time: 1 minute
Teaching Skill: Summarization
Teacher’s Activity: The teacher summarizes the key rules of differentiation discussed and assigns homework.
Pupils’ Activity: Pupils listen to the summary and copy the homework.
Learning Point: Pupils consolidate their learning and prepare for further practice.

Lesson Keywords

• Differentiation – The process of finding the derivative of a function.
• Derivative – The rate at which a function changes with respect to a variable.
• Product Rule – A rule for differentiating the product of two functions.
• Quotient Rule – A rule for differentiating the ratio of two functions.
• Chain Rule – A rule for differentiating composite functions.
• Function – A relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

Differentiation

For pupils who grasp concepts quickly, the teacher will provide more complex problems involving combinations of rules. For pupils needing more support, the teacher will offer one-on-one guidance and simpler, step-by-step examples, encouraging peer tutoring.

Note for teachers using this lesson plan

Ensure pupils have a strong foundation in algebraic manipulation and indices before introducing differentiation. Emphasize the formulas and their correct application through numerous worked examples. Encourage pupils to show all steps in their calculations to minimize errors and aid understanding.