Chapter 2: Differentiation

Overview

Differentiation is a fundamental concept in calculus that deals with rates of change and slopes of curves. This chapter explores the principles of limits, the definition and calculation of derivatives, and various applications of differentiation in mathematics and real-world problems. Mastery of differentiation is essential for understanding advanced mathematical concepts and their applications in physics, engineering, economics, and other sciences.

Learning Objectives

After completing this chapter, you will be able to:

• Understand and evaluate limits
• Define derivatives using the first principle
• Apply differentiation rules and techniques
• Find derivatives of various functions
• Use derivatives to solve optimization and rate of change problems
• Apply differentiation in real-world applications

Key Concepts

2.1 Limit and its Relation to Differentiation

Definition of Limit

The limit of a function f(x) as x approaches a value a is the value that f(x) approaches as x gets arbitrarily close to a.

Notation:

$$\lim_{x \to a} f(x)$$

First Principle of Differentiation

The derivative of a function y = f(x) using the first principle is defined as:

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

Where:

• dy/dx is the derivative of y with respect to x
• h is a small change in x
• f(x + h) - f(x) is the corresponding change in y

Geometric Interpretation

The derivative dy/dx represents:

• The gradient (slope) of the tangent to the curve y = f(x) at point (x, y)
• The rate of change of y with respect to x
• The instantaneous velocity in motion problems

2.2 First and Second Derivatives

First Derivative

The first derivative of y = f(x) is denoted as:

• dy/dx
• f'(x)
• y'

Applications of First Derivative:

• Finding slopes of tangents
• Determining increasing/decreasing functions
• Finding maximum and minimum points

Second Derivative

The second derivative is the derivative of the first derivative:

• $d^2$y/d$x^2$
• f''(x)
• y''

Applications of Second Derivative:

• Determining concavity (concave up/down)
• Finding maximum and minimum points
• Analyzing acceleration in motion

Differentiation Rules

1. Power Rule: If y = ax^n, then dy/dx = anx^(n-1)
2. Product Rule: If y = uv, then dy/dx = u dv/dx + v du/dx
3. Quotient Rule: If y = u/v, then dy/dx = (v du/dx - u dv/dx)/$v^2$
4. Chain Rule: If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx

2.3 Application of Differentiation

Tangent and Normal Lines

Equation of Tangent:

$$y - y_1 = m_T(x - x_1)$$

Where m_T = dy/dx at ($x_1$, $y_1$)

Equation of Normal:

$$y - y_1 = m_N(x - x_1)$$

Where m_N = -1/m_T (negative reciprocal of tangent slope)

Maximum and Minimum Points

First Derivative Test:

• dy/dx > 0: Function is increasing
• dy/dx < 0: Function is decreasing
• dy/dx = 0: Possible maximum, minimum, or point of inflection

Second Derivative Test:

• If dy/dx = 0 and $d^2$y/d$x^2$ > 0: Minimum point
• If dy/dx = 0 and $d^2$y/d$x^2$ < 0: Maximum point

Related Rates of Change

If y = f(x) and x = g(t), then:

$$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$$

This is used to find rates of change when variables are related.

Important Formulas and Methods

Key Differentiation Formulas

Problem-Solving Strategies

Basic Differentiation:

1. Identify the type of function
2. Apply appropriate differentiation rule
3. Simplify the result

Finding Tangents and Normals:

1. Find dy/dx to get gradient function
2. Substitute point to find specific gradient
3. Use point-gradient form for equation

Optimization Problems:

1. Define variables and function
2. Find derivative and set to zero
3. Use second derivative test
4. Check reasonableness of solution

Solved Examples

Example 1: Basic Differentiation

Find the derivatives using first principle: a) f(x) = 3x + 2 b) f(x) = $x^2$

Solutions:

a) f(x) = 3x + 2 f(x + δx) = 3(x + δx) + 2 = 3x + 3δx + 2 δy = f(x + δx) - f(x) = (3x + 3δx + 2) - (3x + 2) = 3δx dy/dx = lim(δx→0) δy/δx = lim(δx→0) 3δx/δx = lim(δx→0) 3 = 3

b) f(x) = $x^2$ f(x + δx) = (x + δx)² = $x^2$ + 2xδx + (δx)² δy = f(x + δx) - f(x) = ($x^2$ + 2xδx + (δx)²) - $x^2$ = 2xδx + (δx)² dy/dx = lim(δx→0) δy/δx = lim(δx→0) (2xδx + (δx)²)/δx = lim(δx→0) (2x + δx) = 2x

Example 2: Differentiation Rules

Find the derivatives: a) y = 4$x^3$ - 2$x^2$ + 5x - 7 b) y = (2x - 3)(3x + 1) c) y = x/(x + 1)

Solutions:

a) Using power rule: dy/dx = 4(3$x^2$) - 2(2x) + 5(1) - 0 = 12$x^2$ - 4x + 5

b) Using product rule (u = 2x - 3, v = 3x + 1): du/dx = 2, dv/dx = 3 dy/dx = u dv/dx + v du/dx = (2x - 3)(3) + (3x + 1)(2) = 6x - 9 + 6x + 2 = 12x - 7

c) Using quotient rule (u = x, v = x + 1): du/dx = 1, dv/dx = 1 dy/dx = (v du/dx - u dv/dx)/$v^2$ = ((x + 1)(1) - (x)(1))/(x + 1)² = (x + 1 - x)/(x + 1)² = 1/(x + 1)²

Example 3: Tangent and Normal Lines

Find the equation of the tangent and normal to the curve y = $x^3$ - 3$x^2$ + 2 at the point (1, 0).

Solution:

First, find dy/dx: y = $x^3$ - 3$x^2$ + 2 dy/dx = 3$x^2$ - 6x

At point (1, 0): m_T = dy/dx = 3(1)² - 6(1) = 3 - 6 = -3

Equation of tangent: y - 0 = -3(x - 1) ⇒ y = -3x + 3

Equation of normal (m_N = -1/m_T = 1/3): y - 0 = (1/3)(x - 1) ⇒ y = (1/3)x - 1/3

Example 4: Maximum and Minimum Points

Find the maximum and minimum points of y = $x^3$ - 3$x^2$ - 9x + 5.

Solution:

Find dy/dx and set to zero: y = $x^3$ - 3$x^2$ - 9x + 5 dy/dx = 3$x^2$ - 6x - 9 Set dy/dx = 0: 3$x^2$ - 6x - 9 = 0 ⇒ $x^2$ - 2x - 3 = 0 (x - 3)(x + 1) = 0 ⇒ x = 3 or x = -1

Find $d^2$y/d$x^2$ for second derivative test: $d^2$y/d$x^2$ = 6x - 6

At x = 3: $d^2$y/d$x^2$ = 6(3) - 6 = 12 > 0 ⇒ Minimum point At x = -1: $d^2$y/d$x^2$ = 6(-1) - 6 = -12 < 0 ⇒ Maximum point

Find y-coordinates: For x = 3: y = 27 - 27 - 27 + 5 = -22 For x = -1: y = -1 - 3 + 9 + 5 = 10

Therefore: Maximum at (-1, 10), Minimum at (3, -22)

Example 5: Related Rates of Change

The radius of a circle is increasing at a rate of 2 cm/s. Find the rate of increase of the area when the radius is 5 cm.

Solution:

Given: dr/dt = 2 cm/s, r = 5 cm Find: dA/dt when r = 5 cm

Area formula: A = π$r^2$ Differentiate both sides with respect to t: dA/dt = 2πr dr/dt

Substitute values: dA/dt = 2π(5)(2) = 20π c$m^2$/s

The area is increasing at 20π c$m^2$/s when the radius is 5 cm.

Mathematical Derivations

Derivation of Power Rule

For y = x^n: Using first principle: f(x + δx) = (x + δx)^n = x^n + nx^(n-1)δx + (n(n-1)/2)x^(n-2)(δx)² + ... + (δx)^n δy = f(x + δx) - f(x) = nx^(n-1)δx + (n(n-1)/2)x^(n-2)(δx)² + ... + (δx)^n dy/dx = lim(δx→0) δy/δx = lim(δx→0) [nx^(n-1) + (n(n-1)/2)x^(n-2)δx + ... + δx^(n-1)] = nx^(n-1)

Derivation of Product Rule

Let y = uv, where u and v are functions of x. Using first principle: y + δy = (u + δu)(v + δv) = uv + uδv + vδu + δuδv δy = uδv + vδu + δuδv δy/δx = u(δv/δx) + v(δu/δx) + δu(δv/δx)

Taking limit as δx → 0: dy/dx = u dv/dx + v du/dx + 0 Therefore: dy/dx = u dv/dx + v du/dx

Derivation of Chain Rule

Let y = f(u) and u = g(x) Using first principle: δy/δx = (δy/δu)(δu/δx) Taking limit as δx → 0: dy/dx = dy/du × du/dx

Real-World Applications

1. Physics - Motion

Velocity and Acceleration:

• Position: s(t)
• Velocity: v = ds/dt
• Acceleration: a = dv/dt = $d^2$s/d$t^2$

Example: A particle moves according to s(t) = $t^3$ - 6$t^2$ + 9t. Find when it is at rest.

2. Economics - Optimization

Marginal Cost and Revenue:

• Total cost: TC(q)
• Marginal cost: MC = d(TC)/dq
• Total revenue: TR(q)
• Marginal revenue: MR = d(TR)/dq

Example: Maximize profit π = TR - TC by finding critical points.

3. Engineering - Structural Analysis

Beam Deflection:

• The deflection of a beam under load involves finding derivatives of the deflection equation.
• Maximum stress occurs where the second derivative is zero.

Example: Find maximum deflection in a simply supported beam.

4. Biology - Population Growth

Rate of Growth:

• Population growth rate: dP/dt
• Exponential growth: P(t) = $P_0$e^(rt)
• Logistic growth: dP/dt = rP(1 - P/K)

Example: Find maximum growth rate in logistic population model.

Complex Problem-Solving Techniques

Problem: Find the coordinates of the points on the curve y = $x^3$ - 6$x^2$ + 9x + 1 where the gradient is 6.

Solution:

Find dy/dx and set equal to 6: y = $x^3$ - 6$x^2$ + 9x + 1 dy/dx = 3$x^2$ - 12x + 9 Set dy/dx = 6: 3$x^2$ - 12x + 9 = 6 ⇒ 3$x^2$ - 12x + 3 = 0 ⇒ $x^2$ - 4x + 1 = 0

Solve quadratic equation: x = [4 ± √(16 - 4)]/2 = [4 ± √12]/2 = [4 ± 2√3]/2 = 2 ± √3

Find y-coordinates: For x = 2 + √3: y = (2 + √3)³ - 6(2 + √3)² + 9(2 + √3) + 1 For x = 2 - √3: y = (2 - √3)³ - 6(2 - √3)² + 9(2 - √3) + 1

After calculation: Points are (2 + √3, 7 + 4√3) and (2 - √3, 7 - 4√3)

Problem: A rectangular box has a square base and volume 32 $m^3$. Find the dimensions that minimize the surface area.

Solution:

Let base side = x, height = h Volume: $x^2$h = 32 ⇒ h = 32/$x^2$

Surface area: S = 2$x^2$ + 4xh = 2$x^2$ + 4x(32/$x^2$) = 2$x^2$ + 128/x

Find dS/dx and set to zero: dS/dx = 4x - 128/$x^2$ Set dS/dx = 0: 4x - 128/$x^2$ = 0 ⇒ 4x = 128/$x^2$ ⇒ 4$x^3$ = 128 ⇒ $x^3$ = 32 ⇒ x = ³√32 = 2³√4

Find h: h = 32/$x^2$ = 32/(4 × 2²√2) = 8/2²√2 = 4 × 2^(1/2) = 4√2

Dimensions: Base 2³√4 m, Height 4√2 m

Problem: Related Rates - A ladder 10 m long rests against a vertical wall. If the bottom slides away at 2 m/s, how fast is the top sliding down when the bottom is 6 m from the wall?

Solution:

Let x = distance from wall, y = height on wall Given: dx/dt = 2 m/s, x = 6 m, ladder = 10 m

From Pythagoras: $x^2$ + $y^2$ = 10² = 100 At x = 6: 36 + $y^2$ = 100 ⇒ $y^2$ = 64 ⇒ y = 8 m

Differentiate with respect to t: 2x dx/dt + 2y dy/dt = 0 Substitute: 2(6)(2) + 2(8) dy/dt = 0 ⇒ 24 + 16 dy/dt = 0 ⇒ dy/dt = -24/16 = -1.5 m/s

The top is sliding down at 1.5 m/s.

Summary Points

• Derivatives represent rates of change and slopes of tangents
• First principle definition: dy/dx = lim(δx→0) [f(x+δx) - f(x)]/δx
• Power rule: d/dx(x^n) = nx^(n-1)
• Product rule: d/dx(uv) = u dv/dx + v du/dx
• Quotient rule: d/dx(u/v) = (v du/dx - u dv/dx)/$v^2$
• Chain rule: d/dx[f(g(x))] = f'(g(x)) × g'(x)
• Applications include optimization, motion, economics, and engineering

Common Mistakes to Avoid

1. Chain rule errors - Don't forget to multiply by the derivative of the inner function
2. Product/quotient rule confusion - Apply the correct rule for the function type
3. Critical point errors - Check both first and second derivatives
4. Related rates errors - Differentiate with respect to time, not the variable
5. Simplification errors - Always simplify derivatives where possible

SPM Exam Tips

Exam Strategies

1. Memorize rules - Power, product, quotient, and chain rules
2. Show working - Step-by-step differentiation for partial marks
3. Practice first principle - Be prepared to use the limit definition
4. Check critical points - Use both first and second derivative tests
5. Draw diagrams - For related rates problems

Key Exam Topics

• Basic differentiation rules (30% of questions)
• Tangent and normal lines (20% of questions)
• Maximum and minimum problems (25% of questions)
• Related rates (15% of questions)
• First principle applications (10% of questions)

Time Management Tips

• Basic differentiation: 3-4 minutes
• Tangent/normal problems: 4-5 minutes
• Optimization problems: 6-8 minutes
• Related rates problems: 7-10 minutes

Practice Problems

Level 1: Basic Differentiation

1. Find derivatives using first principle: a) f(x) = 5x - 3 b) f(x) = $x^2$ + 2x

2. Find derivatives using rules: a) y = 4$x^3$ - 2$x^2$ + 7x - 1 b) y = (3x + 2)(2x - 1) c) y = x/(x + 2)

Level 2: Tangent and Normal

3. Find equations of tangent and normal to: a) y = $x^2$ - 4x + 3 at (2, -1) b) y = $x^3$ - 3x + 1 at (1, -1)

Level 3: Maximum and Minimum

4. Find maximum and minimum points: a) y = $x^3$ - 6$x^2$ + 9x + 2 b) y = 2$x^3$ - 3$x^2$ - 12x + 1

Level 4: Related Rates

5. Physics: A stone thrown vertically has height h(t) = 20t - 5$t^2$. Find maximum height and when it occurs.

6. Geometry: A circular metal plate expands when heated. If radius increases at 0.5 mm/s, find rate of area increase when radius is 50 mm.

7. Economics: Cost function C(x) = $x^3$ - 6$x^2$ + 15x. Find minimum average cost.

Level 5: Applications

8. Optimization: Find the dimensions of a rectangle with area 100 $m^2$ that has minimum perimeter.

9. Motion: A particle moves with position s(t) = $t^3$ - 6$t^2$ + 9t. Find when it changes direction.

10. Related Rates: A ladder 15 ft long leans against a wall. The bottom slides away at 3 ft/s. How fast is the top sliding down when bottom is 9 ft from wall?

Did You Know? 📚

The concept of differentiation was independently developed by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. Newton called it "the method of fluxions" and developed it to solve problems in physics, particularly motion. Leibniz developed a more systematic notation (dy/dx) that we use today. The famous dispute over priority between Newton and Leibniz dominated mathematics for decades and affected the development of calculus in England versus continental Europe.

Quick Reference Guide

---

Differentiation is one of the most powerful tools in mathematics, enabling us to understand and optimize change in countless real-world scenarios. Master these concepts as they form the foundation for integral calculus and advanced mathematical analysis.