Unit 2: Differentiation

Master the definition and basic rules of derivatives

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📋 Unit Overview

Unit 2 introduces the fundamental concept of differentiation. You'll learn how to:

  • Understand the definition of the derivative
  • Calculate derivatives using the limit definition
  • Apply basic differentiation rules
  • Find derivatives of polynomial and power functions
  • Use the product and quotient rules
  • Apply the chain rule for composite functions
💡 Key Insight: The derivative represents the instantaneous rate of change of a function. It tells us how fast a function is changing at any given point, which is crucial for understanding motion, optimization, and many real-world applications.

📐 Definition of the Derivative

The Limit Definition

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

f'(a) = limh→0 [f(a+h) - f(a)]/h

This is the slope of the tangent line at x = a

🎯 Key Concept

The derivative f'(x) gives us the slope of the tangent line to the curve y = f(x) at any point x.

📝 Example 2.1: Using the Definition

Find f'(2) for f(x) = x² using the limit definition.

Solution:

f'(2) = limh→0 [f(2+h) - f(2)]/h

= limh→0 [(2+h)² - 2²]/h

= limh→0 [4 + 4h + h² - 4]/h

= limh→0 [4h + h²]/h

= limh→0 (4 + h) = 4

Alternative Notation

There are several ways to write the derivative:

f'(x)

Prime notation

dy/dx

Leibniz notation

Dxf(x)

Operator notation

d/dx[f(x)]

Differential operator

⚡ Basic Differentiation Rules

Power Rule

For any real number n:

d/dx[xⁿ] = nxⁿ⁻¹

The most fundamental rule

📝 Example 2.2: Power Rule

Find the derivative of f(x) = x⁵

Solution: f'(x) = 5x⁴

Find the derivative of g(x) = x⁻³

Solution: g'(x) = -3x⁻⁴ = -3/x⁴

Constant Multiple Rule

d/dx[cf(x)] = c · f'(x)

Constants can be factored out

📝 Example 2.3: Constant Multiple

Find the derivative of f(x) = 7x³

Solution: f'(x) = 7 · d/dx[x³] = 7 · 3x² = 21x²

Sum and Difference Rules

d/dx[f(x) ± g(x)] = f'(x) ± g'(x)

Derivatives can be added and subtracted

📝 Example 2.4: Sum Rule

Find the derivative of f(x) = x⁴ + 3x² - 5x + 2

Solution:

f'(x) = d/dx[x⁴] + d/dx[3x²] - d/dx[5x] + d/dx[2]

= 4x³ + 6x - 5 + 0

= 4x³ + 6x - 5

✖️➗ Product and Quotient Rules

Product Rule

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

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

📝 Example 2.5: Product Rule

Find the derivative of f(x) = (x² + 1)(x³ - 2x)

Solution:

Let f(x) = x² + 1 and g(x) = x³ - 2x

f'(x) = 2x and g'(x) = 3x² - 2

Using the product rule:

d/dx[f(x) · g(x)] = (2x)(x³ - 2x) + (x² + 1)(3x² - 2)

= 2x⁴ - 4x² + 3x⁴ - 2x² + 3x² - 2

= 5x⁴ - 3x² - 2

Quotient Rule

d/dx[f(x)/g(x)] = [f'(x) · g(x) - f(x) · g'(x)] / [g(x)]²

"Low d-high minus high d-low, over low squared"

📝 Example 2.6: Quotient Rule

Find the derivative of f(x) = (x² + 1)/(x - 1)

Solution:

Let f(x) = x² + 1 and g(x) = x - 1

f'(x) = 2x and g'(x) = 1

Using the quotient rule:

d/dx[f(x)/g(x)] = [(2x)(x - 1) - (x² + 1)(1)] / (x - 1)²

= [2x² - 2x - x² - 1] / (x - 1)²

= (x² - 2x - 1) / (x - 1)²

🔗 Chain Rule

The chain rule is used for composite functions (functions within functions):

d/dx[f(g(x))] = f'(g(x)) · g'(x)

"Derivative of outside times derivative of inside"

🎯 Key Concept

Think of it as: "What's the derivative of the outside function, evaluated at the inside function, times the derivative of the inside function."

📝 Example 2.7: Chain Rule

Find the derivative of f(x) = (x² + 3x)⁵

Solution:

Let g(x) = x² + 3x (inside function)

Let f(u) = u⁵ (outside function)

g'(x) = 2x + 3

f'(u) = 5u⁴

Using the chain rule:

f'(x) = f'(g(x)) · g'(x) = 5(x² + 3x)⁴ · (2x + 3)

⭐ Special Derivatives to Memorize

Function Derivative Notes
sin(x) cos(x) Trigonometric
cos(x) -sin(x) Trigonometric
eˣ eˣ Exponential
ln(x) 1/x Logarithmic
aˣ aˣ ln(a) General exponential
loga(x) 1/(x ln(a)) General logarithm

🧮 Practice Problems

Problem 1

Find the derivative of f(x) = 3x⁴ - 2x³ + 5x - 7

Solution:

f'(x) = d/dx[3x⁴] - d/dx[2x³] + d/dx[5x] - d/dx[7]

= 3 · 4x³ - 2 · 3x² + 5 · 1 - 0

= 12x³ - 6x² + 5

Problem 2

Find the derivative of f(x) = (x² + 1)(x³ - 2)

Solution:

Using the product rule with f(x) = x² + 1 and g(x) = x³ - 2:

f'(x) = 2x and g'(x) = 3x²

d/dx[f(x) · g(x)] = (2x)(x³ - 2) + (x² + 1)(3x²)

= 2x⁴ - 4x + 3x⁴ + 3x²

= 5x⁴ + 3x² - 4x

Problem 3

Find the derivative of f(x) = (2x + 1)⁴

Solution:

Using the chain rule with g(x) = 2x + 1 and f(u) = u⁴:

g'(x) = 2 and f'(u) = 4u³

f'(x) = f'(g(x)) · g'(x) = 4(2x + 1)³ · 2

= 8(2x + 1)³

💡 Exam Tips

🎯 Multiple Choice
  • Always check your work by taking the derivative of your answer
  • Look for patterns in the answer choices
  • Use the power rule for polynomials
  • Remember the chain rule for composite functions
✍️ Free Response
  • Show all steps clearly
  • State which rule you're using
  • Simplify your final answer
  • Check that your answer makes sense
⚠️ Common Mistakes:
  • Forgetting the chain rule for composite functions
  • Mixing up the product and quotient rules
  • Not simplifying the final answer
  • Forgetting to apply the power rule correctly

🚀 Ready to Test Your Knowledge?

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➡️ What's Next?

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In Unit 3, you'll learn about applications of derivatives including optimization and related rates.

Go to Unit 3: Applications of Derivatives →