IB Mathematics AA – Topic 5: Calculus

Comprehensive Guide to Differential Calculus

Introduction to Differential Calculus

Differential calculus is the mathematics of change and rates of change. It provides the tools to analyze how functions behave—where they increase or decrease, where they reach maximum or minimum values, and how fast they're changing at any instant. From optimizing profits in business to calculating instantaneous velocities in physics, differential calculus is fundamental to understanding dynamic systems.

Key concepts: The derivative measures the instantaneous rate of change of a function—geometrically, it's the slope of the tangent line to the curve. Differentiation rules (power, product, quotient, chain rule) enable us to find derivatives efficiently. The first derivative reveals where functions increase or decrease and locate turning points, while the second derivative tells us about concavity and inflection points.

Why calculus matters: Differential calculus transforms static mathematics into dynamic analysis. Engineers use it to optimize designs, economists to find maximum profit and minimum cost, physicists to describe motion and forces, and biologists to model population growth. Understanding limits provides the rigorous foundation for derivatives, while L'Hôpital's Rule offers elegant solutions to indeterminate forms.

In this guide: We'll master all differentiation rules with complete proficiency, analyze curves to find stationary points and determine their nature, solve optimization problems to find maximum and minimum values, tackle related rates problems where multiple quantities change simultaneously, understand limits rigorously, and apply L'Hôpital's Rule to evaluate challenging limits—all essential for IB exam success.

1. Differentiation Rules

Basic Differentiation Rules

Essential Derivative Formulas

Power Rule

\(\frac{d}{dx}(x^n) = nx^{n-1}\)

Works for any real number \(n\)

Constant Multiple Rule

\(\frac{d}{dx}[cf(x)] = c \cdot f'(x)\)

Sum/Difference Rule

\(\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)\)

Standard Functions

\(\frac{d}{dx}(e^x) = e^x\)

\(\frac{d}{dx}(\ln x) = \frac{1}{x}\)

\(\frac{d}{dx}(\sin x) = \cos x\)

\(\frac{d}{dx}(\cos x) = -\sin x\)

\(\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}\)

Advanced Rules

Product, Quotient, and Chain Rules

Product Rule

\(\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)\)

"First times derivative of second, plus second times derivative of first"

Quotient Rule

\(\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\)

"Low dee-high minus high dee-low, all over low squared"

Chain Rule

\(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

Derivative of outside × derivative of inside

Example: \(\frac{d}{dx}[(3x+1)^5] = 5(3x+1)^4 \cdot 3 = 15(3x+1)^4\)

⚠ Differentiation Pitfalls:

Chain rule forgotten: Don't forget to multiply by derivative of inside function!
Product rule misapplication: \((fg)' \neq f'g'\) — you must add both terms
Quotient rule sign: It's minus in numerator (high dee-low minus low dee-high)
Constants vanish: Derivative of constant is zero

2. Properties of Curves

Stationary Points

Finding and Classifying Stationary Points

Stationary Point: Where \(f'(x) = 0\) (tangent is horizontal)

Types of Stationary Points:

Local Maximum: \(f'(x) = 0\) and \(f''(x) < 0\)
Local Minimum: \(f'(x) = 0\) and \(f''(x) > 0\)
Point of Inflection: \(f'(x) = 0\) and \(f''(x) = 0\) (test either side)

First Derivative Test (Sign Analysis)

If \(f'\) changes from + to −: local maximum
If \(f'\) changes from − to +: local minimum
If \(f'\) doesn't change sign: point of inflection

Increasing and Decreasing

Function Behavior:

Increasing: Where \(f'(x) > 0\)
Decreasing: Where \(f'(x) < 0\)
Concave up: Where \(f''(x) > 0\) (curve bends upward)
Concave down: Where \(f''(x) < 0\) (curve bends downward)

Example 1: Properties of Curves

Problem: Given \(f(x) = x^3 - 6x^2 + 9x + 1\)

(a) Find all stationary points

(b) Determine their nature

Solution:

(a) Find stationary points:

Step 1: Find first derivative

\(f'(x) = 3x^2 - 12x + 9\)

Step 2: Set \(f'(x) = 0\)

\(3x^2 - 12x + 9 = 0\)

\(x^2 - 4x + 3 = 0\)

\((x-1)(x-3) = 0\)

\(x = 1\) or \(x = 3\)

Step 3: Find y-coordinates

\(f(1) = 1 - 6 + 9 + 1 = 5\)

\(f(3) = 27 - 54 + 27 + 1 = 1\)

Stationary points: (1, 5) and (3, 1)

(b) Determine nature:

Using second derivative test:

\(f''(x) = 6x - 12\)

At \(x = 1\):

\(f''(1) = 6(1) - 12 = -6 < 0\)

(1, 5) is a local maximum

At \(x = 3\):

\(f''(3) = 6(3) - 12 = 6 > 0\)

(3, 1) is a local minimum

(c) Where function is increasing:

Function increases where \(f'(x) > 0\)

\(3x^2 - 12x + 9 > 0\)

\(3(x-1)(x-3) > 0\)

Testing intervals: positive when \(x < 1\) or \(x > 3\)

Increasing on \((-\infty, 1) \cup (3, \infty)\)

3. Optimization Problems

Optimization Strategy

Steps for Optimization:

Identify the quantity to maximize or minimize
Express this quantity as a function of one variable
Find the derivative and set it equal to zero
Solve for critical points
Use second derivative test or compare values to verify max/min
Check endpoints if domain is restricted

💡 Optimization Tips:

Draw a diagram to visualize the problem
Label all quantities with variables
Eliminate extra variables using constraints
Always check endpoints of domain

Example 2: Optimization Problem (IB-Style)

Problem: A rectangular garden is to be fenced on three sides, with a wall forming the fourth side. If 40 meters of fencing is available, find the dimensions that maximize the area.

Solution:

Step 1: Set up variables

Let \(x\) = width of garden (perpendicular to wall)

Let \(y\) = length of garden (parallel to wall)

Step 2: Express constraint

Fencing needed: \(2x + y = 40\)

Therefore: \(y = 40 - 2x\)

Step 3: Express area in terms of one variable

\(A = xy = x(40 - 2x) = 40x - 2x^2\)

Step 4: Find critical points

\(\frac{dA}{dx} = 40 - 4x\)

Set equal to zero: \(40 - 4x = 0\)

\(x = 10\) meters

Step 5: Verify it's a maximum

\(\frac{d^2A}{dx^2} = -4 < 0\)

Since second derivative is negative, this is a maximum

Step 6: Find dimensions

\(x = 10\) meters

\(y = 40 - 2(10) = 20\) meters

Maximum area: \(A = 10 \times 20 = 200\) m²

Dimensions: 10 m wide × 20 m long, Maximum area: 200 m²

4. Related Rates and Limits

Related Rates

Related Rates Strategy:

Identify all variables and what rates are given/wanted
Write an equation relating the variables
Differentiate both sides with respect to time \(t\) (use chain rule!)
Substitute known values
Solve for the unknown rate

Key: Use \(\frac{d}{dt}\) for all time derivatives

Limits

Limit Definition and Properties

\(\lim_{x \to a} f(x) = L\)

Means: as \(x\) approaches \(a\), \(f(x)\) approaches \(L\)

Limit Laws:

\(\lim[f(x) + g(x)] = \lim f(x) + \lim g(x)\)
\(\lim[f(x) \cdot g(x)] = \lim f(x) \cdot \lim g(x)\)
\(\lim\left[\frac{f(x)}{g(x)}\right] = \frac{\lim f(x)}{\lim g(x)}\) if \(\lim g(x) \neq 0\)

L'Hôpital's Rule

For indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\):

\(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\)

Take derivatives of numerator and denominator separately

Conditions:

Limit must be indeterminate form
Can apply multiple times if needed
Does NOT use quotient rule!

⚠ Limits and L'Hôpital Pitfalls:

L'Hôpital only for indeterminate forms: Check if \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) first!
Not quotient rule: Differentiate top and bottom separately
Can repeat: If still indeterminate, apply L'Hôpital again
Related rates: Differentiate equation first, then substitute values

📋 Differential Calculus Quick Reference

Rule/Concept	Formula	Notes
Power Rule	\(\frac{d}{dx}(x^n) = nx^{n-1}\)	Any real \(n\)
Product Rule	\((fg)' = f'g + fg'\)	Two functions multiplied
Quotient Rule	\(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\)	Low d-hi minus hi d-low
Chain Rule	\(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)	Composite functions
Maximum	\(f'(x) = 0\) and \(f''(x) < 0\)	Or sign change + to −
Minimum	\(f'(x) = 0\) and \(f''(x) > 0\)	Or sign change − to +
L'Hôpital	\(\lim \frac{f}{g} = \lim \frac{f'}{g'}\)	For \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)

🎯 IB Exam Strategy

Common Question Types:

"Find f'(x)": Identify which rule(s) to use, apply systematically
"Find stationary points": Set \(f'(x) = 0\), solve for \(x\), find y-values
"Determine nature": Use second derivative test or first derivative sign test
"Optimize...": Set up function, find derivative, solve, verify max/min
"Evaluate limit": Try substitution first, then L'Hôpital if indeterminate
"Related rates": Write equation, differentiate with respect to time, substitute

Key Reminders:

Chain rule is most commonly forgotten—always check for composition!
Product rule: don't multiply derivatives; add two terms
For stationary points: find where \(f' = 0\) AND calculate y-coordinates
L'Hôpital: only for indeterminate forms, differentiate separately
Optimization: always verify maximum/minimum with second derivative
Related rates: differentiate equation BEFORE substituting values

🎉 Master Differential Calculus!

Differential calculus transforms mathematics from static to dynamic, enabling analysis of change and optimization. From finding maximum profits to understanding instantaneous velocities, derivatives provide the essential tools for modeling the real world. Master differentiation rules, curve analysis, optimization, and limits to excel in IB exams and prepare for advanced calculus and applications across all STEM fields!

Key Success Factors:

✓ Power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\)
✓ Product rule: \((fg)' = f'g + fg'\) (not \(f'g'\)!)
✓ Chain rule: multiply by derivative of inside function
✓ Stationary points: where \(f'(x) = 0\)
✓ Maximum: \(f'' < 0\); Minimum: \(f'' > 0\)
✓ L'Hôpital: for \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), differentiate separately

Apply Rules Correctly • Check Signs • Verify Results

Master differential calculus and excel in IB Mathematics! 🚀