IB Mathematics AI – Topic 5

Notation:

If \(y = f(x)\), the derivative is written as:

\(f'(x)\), \(\frac{dy}{dx}\), or \(y'\)

Power Rule:

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

Works for all \(n \in \mathbb{Q}\)

Standard Derivatives (HL):

• \(\frac{d}{dx}(\sin x) = \cos x\)
• \(\frac{d}{dx}(\cos x) = -\sin x\)
• \(\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}\)
• \(\frac{d}{dx}(e^x) = e^x\)
• \(\frac{d}{dx}(\ln x) = \frac{1}{x}\)

Constant Multiple & Sum Rules:

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

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

⚠️ Common Pitfalls & Tips:

• Power rule: bring exponent down, then reduce exponent by 1
• Constants differentiate to zero
• Use GDC to verify derivatives
• All standard derivatives in formula booklet

Product Rule:

For \(y = u(x) \cdot v(x)\):

\[ \frac{dy}{dx} = u'v + uv' \]

Or in words: "First times derivative of second plus second times derivative of first"

Steps:

1. Identify u and v
2. Find u' and v'
3. Apply formula: u'v + uv'
4. Simplify

Example:

If \(y = x^2 \sin x\):

Let \(u = x^2\), \(v = \sin x\)

Then \(u' = 2x\), \(v' = \cos x\)

\[ \frac{dy}{dx} = 2x \cdot \sin x + x^2 \cdot \cos x \]

⚠️ Common Pitfalls & Tips:

• Don't just multiply derivatives: \((uv)' \neq u'v'\)
• Keep original functions in formula
• Simplify final answer
• Use GDC to verify

Chain Rule:

For \(y = f(g(x))\) or \(y = f(u)\) where \(u = g(x)\):

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

Or: derivative of outer function × derivative of inner function

Steps:

1. Identify outer function f and inner function g
2. Differentiate outer function (keep inner unchanged)
3. Multiply by derivative of inner function

Common Examples:

1. \(y = (3x + 2)^5\)

Outer: \(u^5\), Inner: \(u = 3x + 2\)

\[ \frac{dy}{dx} = 5(3x + 2)^4 \cdot 3 = 15(3x + 2)^4 \]

2. \(y = e^{2x}\)

\[ \frac{dy}{dx} = e^{2x} \cdot 2 = 2e^{2x} \]

3. \(y = \sin(x^2)\)

\[ \frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2) \]

⚠️ Common Pitfalls & Tips:

• CRITICAL: Don't forget to multiply by inner derivative
• Keep inner function unchanged when differentiating outer
• Chain rule needed for ANY composite function
• Practice identifying outer and inner functions

Quotient Rule:

For \(y = \frac{u(x)}{v(x)}\):

\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \]

Memory aid: "Low dee high minus high dee low, all over low squared"

Steps:

1. Identify u (numerator) and v (denominator)
2. Find u' and v'
3. Apply formula: \(\frac{u'v - uv'}{v^2}\)
4. Simplify if possible

Example:

If \(y = \frac{x^2}{x+1}\):

Let \(u = x^2\), \(v = x+1\)

Then \(u' = 2x\), \(v' = 1\)

\[ \frac{dy}{dx} = \frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2} \]

⚠️ Common Pitfalls & Tips:

• Order matters: u'v - uv' NOT uv' - u'v
• Don't forget to square the denominator
• Simplify numerator before final answer
• Alternative: rewrite as product and use product + chain rule

Second Derivative:

The derivative of the derivative:

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

Second Derivative Test:

At a stationary point where \(f'(x_0) = 0\):

• If \(f''(x_0) > 0\): Local MINIMUM (concave up, ∪)
• If \(f''(x_0) < 0\): Local MAXIMUM (concave down, ∩)
• If \(f''(x_0) = 0\): Test inconclusive (use first derivative test)

Concavity:

• \(f''(x) > 0\): Graph is concave up (curves upward)
• \(f''(x) < 0\): Graph is concave down (curves downward)
• \(f''(x) = 0\): Possible point of inflection

Steps to Find & Classify Stationary Points:

1. Find \(f'(x)\)
2. Solve \(f'(x) = 0\) for x-coordinates
3. Find \(f''(x)\)
4. Evaluate \(f''(x_0)\) at each stationary point
5. Classify: positive = min, negative = max

⚠️ Common Pitfalls & Tips:

• Positive second derivative = minimum (think of a smiley face)
• Must have \(f'(x_0) = 0\) first before using second derivative test
• If \(f''(x_0) = 0\), test fails - use first derivative test instead
• Use GDC to verify classification

Solution:

Step 1: Find first derivative

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

Step 2: Solve f'(x) = 0

\[ 3x^2 - 12x + 9 = 0 \]

\[ x^2 - 4x + 3 = 0 \]

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

Stationary points at x = 1 and x = 3

Step 3: Find second derivative

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

Step 4: Classify at x = 1

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

Negative → LOCAL MAXIMUM

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

Maximum at (1, 5)

Step 5: Classify at x = 3

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

Positive → LOCAL MINIMUM

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

Minimum at (3, 1)

Answer: Local maximum at (1, 5); Local minimum at (3, 1)

Tangent Line:

A line that touches the curve at exactly one point

Gradient of tangent = \(f'(x_0)\) at point \((x_0, y_0)\)

Equation of Tangent:

\[ y - y_0 = f'(x_0)(x - x_0) \]

Normal Line:

A line perpendicular to the tangent at that point

Gradient of normal = \(-\frac{1}{f'(x_0)}\) (negative reciprocal)

Equation of Normal:

\[ y - y_0 = -\frac{1}{f'(x_0)}(x - x_0) \]

Steps:

1. Find \(f'(x)\)
2. Evaluate \(f'(x_0)\) at given point
3. Find \(y_0 = f(x_0)\) if not given
4. Use point-gradient form with \((x_0, y_0)\) and gradient
5. Simplify to \(y = mx + c\) form

⚠️ Common Pitfalls & Tips:

• Tangent uses derivative directly
• Normal uses negative reciprocal of derivative
• Must use point-gradient form correctly
• Simplify to y = mx + c form for final answer

Definition:

Finding maximum or minimum values of a function subject to constraints

General Steps:

1. Define variables and express quantity to optimize as function
2. Express any constraints as equations
3. Use constraints to eliminate variables (get single variable function)
4. Differentiate the function
5. Set derivative equal to zero and solve
6. Verify it's max/min using second derivative test or context
7. Answer question in context with units

Common Optimization Problems:

• Maximum area/volume for given perimeter/surface area
• Minimum cost/time/distance
• Maximum profit/revenue
• Optimal dimensions for containers

⚠️ Common Pitfalls & Tips:

• Read question carefully - what are you optimizing?
• Use constraints to reduce to single variable
• Check endpoints if domain is restricted
• Answer in context with correct units

Solution:

Step 1: Define variables

Let x = width (perpendicular to wall)

Let y = length (parallel to wall)

Area to maximize: \(A = xy\)

Step 2: Express constraint

Fencing needed: two widths + one length = 100

\[ 2x + y = 100 \]

\[ y = 100 - 2x \]

Step 3: Express A in terms of x only

\[ A(x) = x(100 - 2x) = 100x - 2x^2 \]

Step 4: Differentiate

\[ \frac{dA}{dx} = 100 - 4x \]

Step 5: Set derivative = 0

\[ 100 - 4x = 0 \]

\[ x = 25 \text{ m} \]

Step 6: Find y

\[ y = 100 - 2(25) = 50 \text{ m} \]

Step 7: Verify maximum

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

Answer: Width = 25m, Length = 50m (Maximum area = 1250 m²)

Definition:

Problems where multiple quantities change with respect to time, and their rates of change are related through an equation

Strategy:

1. Identify all variables and what's changing with time
2. Write equation relating the variables
3. Differentiate both sides with respect to time (use chain rule!)
4. Substitute known values and rates
5. Solve for unknown rate

Key Notation:

• \(\frac{dx}{dt}\) = rate of change of x with respect to time
• Use chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\)

Common Applications:

• Expanding circles/spheres (radius vs. area/volume)
• Filling/emptying tanks
• Moving shadows
• Ladder sliding problems

⚠️ Common Pitfalls & Tips:

• Differentiate equation BEFORE substituting values
• Use chain rule when differentiating with respect to time
• Keep track of units (m/s, cm³/min, etc.)
• Draw diagram to visualize relationships

Rules (HL)

• Product: u'v + uv'
• Quotient: (u'v - uv')/v²
• Chain: outer' × inner'

Second Derivative

• f''(x) > 0: Minimum
• f''(x) < 0: Maximum
• f''(x) = 0: Inconclusive

Lines

• Tangent: m = f'(x₀)
• Normal: m = -1/f'(x₀)
• Use point-gradient form

Optimization

• Set f'(x) = 0
• Solve for x
• Verify max/min