Introductory Calculus: Differentiation

What you'll learn

Differentiation is the mathematical process of finding the rate at which a quantity changes. This topic forms a significant portion of the CSEC Additional Mathematics examination and tests your ability to find derivatives, apply derivative rules, and solve practical problems involving rates of change. You will learn to differentiate polynomial functions, find equations of tangents and normals, and solve optimization problems relevant to Caribbean contexts.

Key terms and definitions

Derivative — the rate of change of a function with respect to its variable; written as dy/dx or f'(x), it represents the gradient of the curve at any point.

Differentiation — the process of finding the derivative of a function.

Gradient function — another name for the derivative; a function that gives the gradient of the original curve at any value of x.

Tangent — a straight line that touches a curve at exactly one point and has the same gradient as the curve at that point.

Normal — a straight line perpendicular to the tangent at a given point on a curve.

Stationary point — a point on a curve where the gradient equals zero (dy/dx = 0); the curve is momentarily neither increasing nor decreasing.

Rate of change — how quickly one quantity changes with respect to another; differentiation measures instantaneous rate of change.

Power rule — the fundamental rule for differentiating terms of the form ax^n, giving nax^(n-1).

Core concepts

The derivative from first principles

The derivative of a function y = f(x) is defined as:

dy/dx = lim[h→0] [f(x+h) - f(x)]/h

While you may need to recognize this definition, CSEC examinations typically focus on applying differentiation rules rather than deriving them from first principles. The definition establishes that the derivative represents the limit of the gradient of a chord as the two points become infinitesimally close.

Basic differentiation rules

The Power Rule

For any function y = ax^n where a and n are constants:

dy/dx = nax^(n-1)

This is the most frequently used rule at CSEC level. Apply it term by term to polynomials.

Examples:

• If y = x^5, then dy/dx = 5x^4
• If y = 4x^3, then dy/dx = 12x^2
• If y = 7x^2, then dy/dx = 14x
• If y = x, then dy/dx = 1
• If y = 6 (constant), then dy/dx = 0

Differentiating sums and differences

Differentiate each term separately:

If y = f(x) + g(x), then dy/dx = f'(x) + g'(x)

Example: If y = 3x^4 - 5x^2 + 2x - 7, then dy/dx = 12x^3 - 10x + 2

Negative and fractional powers

Express roots and fractions as powers before differentiating:

• √x = x^(1/2), so d/dx(√x) = (1/2)x^(-1/2) = 1/(2√x)
• 1/x^2 = x^(-2), so d/dx(x^(-2)) = -2x^(-3) = -2/x^3

Finding gradients at specific points

To find the gradient of a curve at a particular point:

1. Differentiate the function to find dy/dx
2. Substitute the x-coordinate into dy/dx
3. Calculate the numerical value

Example: For y = x^3 - 4x + 1, find the gradient when x = 2.

dy/dx = 3x^2 - 4

At x = 2: gradient = 3(2)^2 - 4 = 12 - 4 = 8

Equations of tangents and normals

Tangent lines

The equation of a tangent to a curve at point (x₁, y₁):

1. Find dy/dx and evaluate at x = x₁ to get the gradient m
2. Use the point-slope form: y - y₁ = m(x - x₁)
3. Rearrange to y = mx + c form if required

Normal lines

The normal is perpendicular to the tangent. If the tangent has gradient m, the normal has gradient -1/m.

Process:

1. Find the tangent gradient m at the point
2. Normal gradient = -1/m
3. Use y - y₁ = (-1/m)(x - x₁)

Stationary points and curve sketching

Stationary points occur where dy/dx = 0. To locate them:

1. Differentiate the function
2. Set dy/dx = 0
3. Solve for x
4. Substitute x-values back into the original equation to find y-coordinates

Stationary points can be:

• Maximum points — gradient changes from positive to negative
• Minimum points — gradient changes from negative to positive
• Points of inflection — gradient remains positive or negative on both sides

At CSEC level, you identify the nature of stationary points by testing the sign of dy/dx on either side of the stationary point, or by sketching the curve.

Applications: rates of change

Differentiation solves real-world problems involving rates of change. Caribbean contexts include:

Population growth: If a model gives the population P of green vervet monkeys in Barbados after t years, dP/dt represents the rate of population change.

Economic applications: If C(x) represents the cost in EC dollars to produce x bottles of hot pepper sauce, dC/dx gives the marginal cost (additional cost to produce one more bottle).

Motion: For displacement s at time t, ds/dt gives velocity and d²s/dt² gives acceleration.

Problem-solving steps:

1. Identify what is changing and with respect to what
2. Write the relationship as a function
3. Differentiate with respect to the appropriate variable
4. Substitute given values

Connected rates of change

When two variables both depend on a third variable (often time), use the chain rule relationship:

dy/dt = (dy/dx) × (dx/dt)

This appears in problems where multiple quantities change simultaneously.

Example: The radius of a circular oil spill from a tanker near Trinidad increases at 2 m/s. Find how fast the area is increasing when r = 50 m.

Given: dr/dt = 2 m/s Area: A = πr² dA/dr = 2πr dA/dt = (dA/dr) × (dr/dt) = 2πr × 2 = 4πr

When r = 50: dA/dt = 4π(50) = 200π ≈ 628.3 m²/s

Worked examples

Example 1: Basic differentiation and gradient

Question: The curve C has equation y = 2x³ - 5x² + 3x - 1.

(a) Find dy/dx. [2 marks] (b) Calculate the gradient of C at the point where x = -1. [2 marks] (c) Determine the coordinates of the points on C where the gradient is 7. [4 marks]

Solution:

(a) y = 2x³ - 5x² + 3x - 1

Differentiating term by term using the power rule: dy/dx = 6x² - 10x + 3 ✓ [2 marks]

(b) At x = -1: dy/dx = 6(-1)² - 10(-1) + 3 ✓ = 6 + 10 + 3 = 19 ✓ [2 marks]

(c) When gradient = 7: 6x² - 10x + 3 = 7 ✓ 6x² - 10x - 4 = 0 3x² - 5x - 2 = 0 (3x + 1)(x - 2) = 0 ✓ x = -1/3 or x = 2 ✓

When x = -1/3: y = 2(-1/3)³ - 5(-1/3)² + 3(-1/3) - 1 = -70/27 When x = 2: y = 2(8) - 5(4) + 3(2) - 1 = 1

Points: (-1/3, -70/27) and (2, 1) ✓ [4 marks]

Example 2: Tangent and normal equations

Question: A curve has equation y = x² - 6x + 5.

(a) Find the equation of the tangent to the curve at the point (1, 0). [4 marks] (b) Find the equation of the normal at this point. [3 marks]

Solution:

(a) y = x² - 6x + 5 dy/dx = 2x - 6 ✓

At x = 1: gradient of tangent = 2(1) - 6 = -4 ✓

Using y - y₁ = m(x - x₁): y - 0 = -4(x - 1) ✓ y = -4x + 4 ✓ [4 marks]

(b) Gradient of normal = -1/(-4) = 1/4 ✓

y - 0 = (1/4)(x - 1) ✓ y = (1/4)x - 1/4 or 4y = x - 1 ✓ [3 marks]

Example 3: Application problem

Question: A rectangular plot of land in Jamaica is to be fenced on three sides, with an existing wall forming the fourth side. The farmer has 240 metres of fencing available. If the width of the plot perpendicular to the wall is x metres, show that the area A = 240x - 2x². Find the maximum possible area. [8 marks]

Solution:

Let width = x metres (perpendicular to wall) Let length = y metres (parallel to wall)

Total fencing: 2x + y = 240 ✓ Therefore: y = 240 - 2x ✓

Area: A = xy = x(240 - 2x) A = 240x - 2x² ✓ [as required]

For maximum area, dA/dx = 0: dA/dx = 240 - 4x ✓

Setting equal to zero: 240 - 4x = 0 ✓ x = 60 ✓

When x = 60: A = 240(60) - 2(60)² A = 14400 - 7200 A = 7200 m² ✓ [8 marks]

Common mistakes and how to avoid them

• Forgetting to reduce the power by 1: When differentiating x^n, students often multiply by n but forget to write x^(n-1). Always apply both steps of the power rule.

• Incorrect handling of negative powers: When differentiating terms like 1/x², first rewrite as x^(-2), then apply the power rule to get -2x^(-3), not positive. The negative sign in the power creates another negative.

• Mixing up tangent and normal gradients: Remember that if the tangent gradient is m, the normal gradient is -1/m (negative reciprocal), not just the reciprocal or negative.

• Setting the original function to zero instead of the derivative: To find stationary points, set dy/dx = 0, not y = 0. You're looking for where the gradient is zero, not where the curve crosses the x-axis.

• Forgetting to find both coordinates: Questions asking for "the point" require both x and y coordinates. After solving for x, always substitute back into the original equation (not the derivative) to find y.

• Arithmetic errors with substitution: When substituting negative values, use brackets: if x = -2 and you need x², write (-2)² = 4, not -2² = -4.

Exam technique for Introductory Calculus: Differentiation

• "Find dy/dx" or "Differentiate": These commands require you to apply differentiation rules and write the derivative clearly. Show each step for partial credit. Worth 2-3 marks typically, depending on complexity.

• "Hence" or "Show that": When a question asks you to show a given result, you must include all working steps. Simply stating the answer earns no marks. These questions test whether you can derive the result independently, often worth 2-3 marks.

• Multi-part questions build on each other: Your answer to part (a) is usually needed for parts (b) and (c). If you cannot complete part (a), use the expected result to continue—you can still earn method marks in subsequent parts.

• Application problems require units: When finding rates of change in context (speed, cost, population growth), always include appropriate units in your final answer. State what the answer represents: "The population is increasing at 125 monkeys per year when t = 5."

Quick revision summary

Differentiation finds the rate of change or gradient of a function. Apply the power rule: d/dx(ax^n) = nax^(n-1). Differentiate polynomials term by term. Find gradients at specific points by substituting into dy/dx. Tangents touch the curve with gradient m; normals are perpendicular with gradient -1/m. Stationary points occur where dy/dx = 0. Solve application problems by identifying variables, writing functional relationships, differentiating, then substituting given values. Always show clear working and include units in real-world contexts.