Class: Senior Secondary School 2 (SS2 / SSS2)
Term: Second Term
Week: 5
Age: 16 years
Duration: 45 minutes
Subject: Further Mathematics
Curriculum Theme: Calculus
Previous Lesson: Differentiation: Applications to Rate, Gradient and Optimization.
Topic: DIFFERENTIATION
Subject Matter: Higher derivatives, differentiation of implicit functions

Specific Objectives

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

Cognitive Domain:

Define higher derivatives and implicit functions.
State the rules for finding second and higher derivatives.
Explain the method for differentiating implicit functions.

Affective Domain:

Appreciate the importance of higher derivatives in various applications.
Show interest in solving problems involving implicit differentiation.
Develop a systematic approach to solving calculus problems.

Psychomotor Domain:

Compute the second and higher derivatives of given functions.
Apply the chain rule effectively to differentiate implicit functions.
Accurately solve problems involving both higher derivatives and implicit differentiation.

Participate actively in class discussions and problem-solving sessions.
Collaborate with peers to clarify understanding of differentiation concepts.

Reference Materials

The following resources were used in planning this lesson:

Senior Secondary Schools Education Curriculum
State Unified Scheme of Work
New Further Mathematics for Senior Secondary Schools by P.N. Okeke

Instructional Materials

The teacher will teach this lesson with the aid of:

Chart showing areas of application of differentiation.
Whiteboard and markers.
Textbooks and notebooks.

Rationale for the Lesson

This lesson helps pupils understand how to find derivatives beyond the first order and how to differentiate functions where the dependent variable is not explicitly expressed. These skills are important for solving more complex problems in mathematics, physics, and engineering.

Prerequisite/Previous Knowledge

Pupils have prior knowledge of basic differentiation rules, including the power rule, product rule, quotient rule, and chain rule for explicit functions.

Lesson Content/Board Summary

DIFFERENTIATION

Higher Derivatives

Higher derivatives refer to derivatives of a function beyond the first derivative. The second derivative is the derivative of the first derivative, the third derivative is the derivative of the second derivative, and so on.

Notation for higher derivatives:

First derivative: dy/dx or f'(x)
Second derivative: d²y/dx² or f”(x)
Third derivative: d³y/dx³ or f”'(x)
Nth derivative: dⁿy/dxⁿ or fⁿ(x)

Example 1: Find the second derivative of y = 3x⁴ – 2x³ + 5x – 1.

Solution:
Step 1: Find the first derivative (dy/dx).
dy/dx = d/dx (3x⁴ – 2x³ + 5x – 1)
dy/dx = 12x³ – 6x² + 5

Step 2: Find the second derivative (d²y/dx²) by differentiating dy/dx.
d²y/dx² = d/dx (12x³ – 6x² + 5)
d²y/dx² = 36x² – 12x

Example 2: If f(x) = sin(2x), find f”(x).

Solution:
Step 1: Find the first derivative f'(x).
f'(x) = d/dx (sin(2x))
f'(x) = 2cos(2x)

Step 2: Find the second derivative f”(x) by differentiating f'(x).
f”(x) = d/dx (2cos(2x))
f”(x) = 2 * (-sin(2x) * 2)
f”(x) = -4sin(2x)

Differentiation of Implicit Functions

An implicit function is a function where the dependent variable (usually y) is not explicitly expressed in terms of the independent variable (usually x). For example, x² + y² = 25 is an implicit function.

To differentiate an implicit function, we differentiate both sides of the equation with respect to x, remembering to apply the chain rule whenever we differentiate a term involving y. When differentiating a term with y, we treat y as a function of x, so d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx).

Example 1: Find dy/dx for the implicit function x² + y² = 25.

Solution:
Step 1: Differentiate both sides with respect to x.
d/dx (x² + y²) = d/dx (25)
d/dx (x²) + d/dx (y²) = d/dx (25)

Step 2: Differentiate each term. Remember to use the chain rule for y².
2x + 2y (dy/dx) = 0

Step 3: Isolate dy/dx.
2y (dy/dx) = -2x
dy/dx = -2x / 2y
dy/dx = -x/y

Example 2: Find dy/dx for the implicit function 3xy – y³ = x².

Solution:
Step 1: Differentiate both sides with respect to x. Remember to use the product rule for 3xy.
d/dx (3xy) – d/dx (y³) = d/dx (x²)

Step 2: Differentiate each term.
For 3xy (product rule: u=3x, v=y): u’v + uv’ = 3(y) + 3x(dy/dx)
For y³: 3y²(dy/dx)
For x²: 2x

So, the equation becomes:
3y + 3x(dy/dx) – 3y²(dy/dx) = 2x

Step 3: Group terms containing dy/dx and isolate dy/dx.
3x(dy/dx) – 3y²(dy/dx) = 2x – 3y
dy/dx (3x – 3y²) = 2x – 3y
dy/dx = (2x – 3y) / (3x – 3y²)

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 reviews the concept of the first derivative and asks pupils to recall basic differentiation rules. The teacher then introduces the idea of finding derivatives of derivatives.
Pupils’ Activity: Pupils respond to questions about first derivatives and pay attention to the introduction of higher derivatives.
Learning Point: Pupils recall prior knowledge of differentiation and are introduced to the lesson’s topic.

Step 2: Presentation of Higher Derivatives

Time: 10 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains what higher derivatives are (second, third, etc.) and their standard notations. The teacher demonstrates how to find the second derivative using a simple polynomial function.
Pupils’ Activity: Pupils listen attentively, take notes, and ask questions for clarification on notations and procedures.
Learning Point: Pupils understand the concept and notation of higher derivatives.

Step 3: Worked Examples on Higher Derivatives

Time: 8 minutes
Teaching Skill: Problem Solving/Demonstration
Teacher’s Activity: The teacher works through more complex examples of finding higher derivatives, including trigonometric or exponential functions, guiding pupils through each step.
Pupils’ Activity: Pupils observe the step-by-step solutions, copy the examples, and attempt to solve parts of the problems as guided by the teacher.
Learning Point: Pupils gain practical experience in computing higher derivatives.

Step 4: Introduction to Implicit Functions

Time: 8 minutes
Teaching Skill: Explanation/Concept Building
Teacher’s Activity: The teacher defines implicit functions and contrasts them with explicit functions. The teacher explains the general principle of differentiating implicit functions, emphasizing the application of the chain rule to terms involving ‘y’.
Pupils’ Activity: Pupils listen to the explanation, identify the difference between implicit and explicit functions, and understand the need for the chain rule.
Learning Point: Pupils comprehend the definition and method for differentiating implicit functions.

Step 5: Worked Examples on Implicit Functions

Time: 8 minutes
Teaching Skill: Problem Solving/Demonstration
Teacher’s Activity: The teacher solves at least two examples of implicit differentiation on the board, clearly showing the application of the chain rule and how to isolate dy/dx.
Pupils’ Activity: Pupils follow the worked examples, ask questions about specific steps, and ensure they understand how to apply the chain rule for ‘y’ terms.
Learning Point: Pupils learn to apply the rules of implicit differentiation to solve problems.

Step 6: Evaluation/Review

Time: 5 minutes
Teaching Skill: Questioning/Assessment
Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

What is the second derivative of a function?
Find the second derivative of y = x⁵ – 3x² + 7.
Explain the main difference between an explicit and an implicit function.
Differentiate implicitly to find dy/dx if x³ + y³ = 6xy.

Pupils’ Activity: Pupils answer orally and in writing.
Learning Point: Pupils demonstrate understanding of the lesson.

Step 7: Conclusion

Time: 1 minute
Teaching Skill: Summarization
Teacher’s Activity: The teacher briefly summarizes the key concepts covered: higher derivatives and implicit differentiation. The teacher assigns homework for further practice.
Pupils’ Activity: Pupils listen to the summary and copy down the homework assignment.
Learning Point: Pupils consolidate their understanding and are given tasks to reinforce learning.

Lesson Keywords

Higher Derivatives – Derivatives of a function beyond the first derivative.
Second Derivative – The derivative of the first derivative.
Implicit Function – A function where the dependent variable is not explicitly expressed in terms of the independent variable.
Explicit Function – A function where the dependent variable is isolated on one side of the equation.
Chain Rule – A rule used when differentiating composite functions, crucial for implicit differentiation.

Differentiation

For pupils who grasp concepts quickly, the teacher can provide more challenging problems involving combinations of product/quotient rule with implicit differentiation or higher-order derivatives of complex functions. For pupils who struggle, the teacher will provide additional simpler examples and one-on-one guidance, focusing on the step-by-step application of rules.

Note for teachers using this lesson plan

Ensure pupils have a solid foundation in basic differentiation rules before introducing higher derivatives and implicit functions. Emphasize the systematic application of the chain rule during implicit differentiation. Encourage pupils to show all working steps clearly, especially in the examples, to help them understand the process.