IB Maths AA Analysis and Approaches How to Solve: IB AA HL/SL – Limits and Derivatives (First Principles, Chain Rule, Implicit)

How to Solve: IB AA HL/SL – Limits and Derivatives (First Principles, Chain Rule, Implicit)

Complete Guide

Introduction

Mastering limits and derivatives unlocks 7–10% of your IB AA HL/SL exam score—and powers real-world models in physics (motion), chemistry (reaction rates), biology (population growth), and economics (marginal cost). One slip in the chain rule or first principles, and you lose 4–6 marks on a single question. Let’s fix that.

WHAT YOU NEED TO KNOW FIRST

1. Basic algebra: Expand brackets, factorise quadratics, simplify fractions.
2. Function notation: Understand f(x), f(a + h), and Δx.
3. Limits basics: Know that limₕ→₀ (h) = 0 and how to substitute h = 0 (if no division by zero).

KEY TERMS & FORMULAS

1. First Principles (Definition of the Derivative)

Formula: f'(x) = limₕ→₀ [f(x + h) – f(x)] / h

• f'(x): Derivative of f(x) (slope of the tangent line).
• h: Small change in x (approaches 0).
• MEMORISE THIS: You’ll use it to prove derivatives or when asked for "from first principles."

2. Chain Rule (Composite Functions)

Formula: d/dx [f(g(x))] = f'(g(x)) · g'(x)

• f(g(x)): Outer function f applied to inner function g.
• f'(g(x)): Derivative of f evaluated at g(x).
• g'(x): Derivative of g.
• MEMORISE THIS: Used for nested functions (e.g., sin(3x²), e^(5x+1)).

3. Implicit Differentiation

Formula: Differentiate both sides of the equation with respect to x, treating y as a function of x (use dy/dx for y terms).

• Example: For x² + y² = 25, differentiate to get 2x + 2y(dy/dx) = 0.
• MEMORISE THIS: Used when y is not isolated (e.g., circles, ellipses).

4. Standard Derivatives (Given on IB Exam Sheet)

• d/dx [xⁿ] = nxⁿ⁻¹
• d/dx [sin(x)] = cos(x)
• d/dx [cos(x)] = –sin(x)
• d/dx [eˣ] = eˣ
• d/dx [ln(x)] = 1/x

STEP-BY-STEP METHOD

A. First Principles (Proving a Derivative)

Steps: 1. Write the definition: f'(x) = limₕ→₀ [f(x + h) – f(x)] / h. 2. Substitute f(x + h) and f(x) into the numerator. 3. Expand and simplify the numerator (cancel f(x) terms). 4. Factor out h from the numerator. 5. Cancel h in numerator and denominator. 6. Take the limit as h → 0 (substitute h = 0). 7. Write the final derivative.

Worked Example: Find f'(x) for f(x) = x² from first principles. 1. f'(x) = limₕ→₀ [(x + h)² – x²] / h 2. f'(x) = limₕ→₀ [x² + 2xh + h² – x²] / h 3. f'(x) = limₕ→₀ [2xh + h²] / h 4. f'(x) = limₕ→₀ h(2x + h) / h 5. f'(x) = limₕ→₀ (2x + h) 6. f'(x) = 2x + 0 = 2x

What we did and why: We expanded (x + h)², simplified, and took the limit to prove the derivative of x² is 2x.

B. Chain Rule (Composite Functions)

Steps: 1. Identify the outer function f(u) and inner function u = g(x). 2. Differentiate the outer function f(u) with respect to u: f'(u). 3. Differentiate the inner function g(x) with respect to x: g'(x). 4. Multiply the results: f'(g(x)) · g'(x). 5. Substitute back u = g(x) if needed.

Worked Example: Find dy/dx for y = sin(3x²). 1. Outer: f(u) = sin(u), Inner: u = 3x². 2. f'(u) = cos(u). 3. g'(x) = 6x. 4. dy/dx = cos(3x²) · 6x = 6x cos(3x²).

What we did and why: We treated 3x² as a single variable u, differentiated sin(u), then multiplied by the derivative of 3x².

C. Implicit Differentiation

Steps: 1. Differentiate both sides of the equation with respect to x. 2. For y terms, multiply by dy/dx (chain rule). 3. Collect all dy/dx terms on one side. 4. Factor out dy/dx. 5. Solve for dy/dx.

Worked Example: Find dy/dx for x² + y² = 25. 1. d/dx [x²] + d/dx [y²] = d/dx [25] 2. 2x + 2y(dy/dx) = 0 3. 2y(dy/dx) = –2x 4. dy/dx = –2x / (2y) = –x/y

What we did and why: We treated y as a function of x, used the chain rule on y², and solved for dy/dx.

WORKED EXAMPLES

Example 1 – Basic (First Principles)

Question: Find f'(x) for f(x) = 3x + 2 from first principles. Solution: 1. f'(x) = limₕ→₀ [3(x + h) + 2 – (3x + 2)] / h 2. f'(x) = limₕ→₀ [3x + 3h + 2 – 3x – 2] / h 3. f'(x) = limₕ→₀ [3h] / h = 3

What we did and why: The derivative of a linear function is its slope (3), proven by simplifying the limit.

Example 2 – Medium (Chain Rule)

Question: Find dy/dx for y = e^(4x – 1). Solution: 1. Outer: f(u) = eᵘ, Inner: u = 4x – 1. 2. f'(u) = eᵘ. 3. g'(x) = 4. 4. dy/dx = e^(4x – 1) · 4 = 4e^(4x – 1).

What we did and why: We used the chain rule to differentiate the exponential function with a linear exponent.

Example 3 – Exam-Style (Implicit + Chain Rule)

Question: Find dy/dx for x sin(y) + y² = x at the point (1, 0). Solution: 1. Differentiate both sides:
d/dx [x sin(y)] + d/dx [y²] = d/dx [x] 2. Use product rule on x sin(y):
sin(y) + x cos(y)(dy/dx) + 2y(dy/dx) = 1 3. Collect dy/dx terms:
x cos(y)(dy/dx) + 2y(dy/dx) = 1 – sin(y) 4. Factor dy/dx:
dy/dx [x cos(y) + 2y] = 1 – sin(y) 5. Solve for dy/dx:
dy/dx = (1 – sin(y)) / (x cos(y) + 2y) 6. Substitute (1, 0):
dy/dx = (1 – 0) / (1 · 1 + 0) = 1

What we did and why: We combined the product rule and implicit differentiation, then substituted the point to find the slope.

COMMON MISTAKES

1. MISTAKE: Forgetting to multiply by dy/dx in implicit differentiation.
   WHY IT HAPPENS: Treating y as a constant.
   CORRECT APPROACH: Always use d/dx [y] = dy/dx.

2. MISTAKE: Misapplying the chain rule (e.g., d/dx [sin(3x)] = cos(3x)).
   WHY IT HAPPENS: Forgetting the inner derivative (3).
   CORRECT APPROACH: Multiply by g'(x) (e.g., 3 cos(3x)).

3. MISTAKE: Expanding (x + h)² incorrectly in first principles.
   WHY IT HAPPENS: Algebra errors.
   CORRECT APPROACH: Use (a + b)² = a² + 2ab + b².

4. MISTAKE: Canceling h before factoring in first principles.
   WHY IT HAPPENS: Skipping steps.
   CORRECT APPROACH: Factor h first, then cancel.

5. MISTAKE: Not substituting h = 0 at the end of first principles.
   WHY IT HAPPENS: Forgetting the limit.
   CORRECT APPROACH: Always take limₕ→₀ at the end.

EXAM TRAPS

1. TRAP: Asking for "from first principles" but hiding it in a word problem.
   HOW TO SPOT IT: Look for phrases like "show that" or "prove the derivative."
   HOW TO AVOID IT: Write the definition f'(x) = limₕ→₀ [...] first.

2. TRAP: Giving a function like y = (2x + 1)⁵ and expecting the chain rule, but students use the power rule alone.
   HOW TO SPOT IT: Nested functions (e.g., (...), sin(...)) always need the chain rule.
   HOW TO AVOID IT: Identify the inner and outer functions before differentiating.

3. TRAP: Implicit differentiation with x and y on both sides (e.g., x²y + y³ = x).
   HOW TO SPOT IT: Equations where y isn’t isolated.
   HOW TO AVOID IT: Differentiate term by term, using the product rule where needed.

1-MINUTE RECAP

"Listen up—this is your 60-second survival guide. For first principles, write the limit definition, expand, simplify, and take the limit. For the chain rule, find the outer and inner functions, differentiate both, and multiply. For implicit differentiation, differentiate both sides, multiply by dy/dx for y terms, and solve for dy/dx. Common mistakes? Forgetting the inner derivative in the chain rule, or not multiplying by dy/dx in implicit. Exam traps? Watch for ‘prove the derivative’ (first principles) or nested functions (chain rule). You’ve got this—now go ace that exam!