2.5 Basic Differentiation Rules Homework Answer Key Pdf

2.5 Basic Differentiation Rules: Homework Answer Key (PDF Alternative)

This comprehensive guide serves as a robust alternative to a PDF answer key for the 2.5 Basic Differentiation Rules homework assignment. We'll cover the fundamental rules, provide explanations, and work through example problems to solidify your understanding. This isn't just about providing answers; it's about mastering the concepts behind differentiation.

Understanding the Foundation: Core Differentiation Rules

Before diving into problems, let's review the essential rules that form the bedrock of differential calculus. A strong grasp of these principles is crucial for tackling more complex derivatives later on.

1. The Power Rule:

This is arguably the most fundamental rule. It states:

d/dx (xⁿ) = nxⁿ⁻¹

Where 'n' is any real number.

• Example: Find the derivative of f(x) = x³.

  Applying the power rule: f'(x) = 3x²

• Example (with a negative exponent): Find the derivative of f(x) = 1/x². Rewrite as f(x) = x⁻².

  Applying the power rule: f'(x) = -2x⁻³ = -2/x³

• Example (with a fractional exponent): Find the derivative of f(x) = √x. Rewrite as f(x) = x¹⁄².

  Applying the power rule: f'(x) = (1/2)x⁻¹⁄² = 1/(2√x)

2. The Constant Multiple Rule:

This rule simplifies derivatives involving constant multipliers:

d/dx [cf(x)] = c * d/dx [f(x)]

Where 'c' is a constant.

• Example: Find the derivative of f(x) = 5x².

  Applying the constant multiple rule and then the power rule: f'(x) = 5 * d/dx(x²) = 5 * 2x = 10x

3. The Sum/Difference Rule:

This rule allows us to differentiate functions that are sums or differences of other functions:

d/dx [f(x) ± g(x)] = d/dx [f(x)] ± d/dx [g(x)]

• Example: Find the derivative of f(x) = 3x² + 2x - 7.

  Applying the sum/difference rule and then the power rule and constant rule: f'(x) = d/dx(3x²) + d/dx(2x) - d/dx(7) = 6x + 2 - 0 = 6x + 2

4. The Product Rule:

This rule is essential when dealing with functions that are products of other functions:

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

• Example: Find the derivative of f(x) = (x² + 1)(x - 2). Let f(x) = x² + 1 and g(x) = x - 2. Then f'(x) = 2x and g'(x) = 1.

  Applying the product rule: f'(x) = (x² + 1)(1) + (x - 2)(2x) = x² + 1 + 2x² - 4x = 3x² - 4x + 1

5. The Quotient Rule:

This rule handles derivatives of functions that are quotients of other functions:

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

• Example: Find the derivative of f(x) = (x² + 1) / (x - 2). Let f(x) = x² + 1 and g(x) = x - 2. Then f'(x) = 2x and g'(x) = 1.

  Applying the quotient rule: f'(x) = [(x - 2)(2x) - (x² + 1)(1)] / (x - 2)² = (2x² - 4x - x² - 1) / (x - 2)² = (x² - 4x - 1) / (x - 2)²

Working Through Example Problems: A Step-by-Step Approach

Let's solidify our understanding by working through several problems. Each problem will demonstrate the application of one or more of the rules we've discussed. Remember, practice is key to mastering differentiation.

Problem 1: Find the derivative of f(x) = 4x⁵ - 7x³ + 2x - 9.

Solution:

Apply the sum/difference rule and the power rule:

f'(x) = 4(5x⁴) - 7(3x²) + 2(1) - 0 = 20x⁴ - 21x² + 2

Problem 2: Find the derivative of f(x) = (3x² + 5)(2x - 1).

Solution:

Apply the product rule:

Let g(x) = 3x² + 5 and h(x) = 2x - 1. Then g'(x) = 6x and h'(x) = 2

f'(x) = (3x² + 5)(2) + (2x - 1)(6x) = 6x² + 10 + 12x² - 6x = 18x² - 6x + 10

Problem 3: Find the derivative of f(x) = (x³ + 2x) / (x² - 1).

Solution:

Apply the quotient rule:

Let g(x) = x³ + 2x and h(x) = x² - 1. Then g'(x) = 3x² + 2 and h'(x) = 2x.

f'(x) = [(x² - 1)(3x² + 2) - (x³ + 2x)(2x)] / (x² - 1)² = (3x⁴ + 2x² - 3x² - 2 - 2x⁴ - 4x²) / (x² - 1)² = (x⁴ - 5x² - 2) / (x² - 1)²

Problem 4: Find the derivative of f(x) = 6√x + 4/x³

Solution:

Rewrite the function using exponents: f(x) = 6x¹⁄² + 4x⁻³

Apply the sum rule and power rule:

f'(x) = 6(1/2)x⁻¹⁄² + 4(-3)x⁻⁴ = 3x⁻¹⁄² - 12x⁻⁴ = 3/√x - 12/x⁴

Problem 5: Find the derivative of f(x) = (x² + 3x) (2x⁻¹ + 5x⁻²)

Solution: This problem combines product and power rules.

Let g(x) = x² + 3x and h(x) = 2x⁻¹ + 5x⁻²

g'(x) = 2x + 3

h'(x) = -2x⁻² -10x⁻³

Using the product rule:

f'(x) = (x² + 3x)(-2x⁻² - 10x⁻³) + (2x⁻¹ + 5x⁻²)(2x + 3)

This simplifies to a somewhat lengthy expression, but the core process remains the application of the product and power rules. Expanding and simplifying this expression is left as an exercise to reinforce your understanding of algebraic manipulation.

Beyond the Basics: Expanding Your Differentiation Skills

While these five rules form the core of basic differentiation, there are many other techniques and rules you'll encounter as you progress in calculus. These include:

• Chain Rule: Used to differentiate composite functions (functions within functions).
• Implicit Differentiation: Used to differentiate equations that are not explicitly solved for y.
• Logarithmic Differentiation: Used to differentiate functions that involve products, quotients, and powers of other functions.
• Trigonometric Derivatives: Covers the derivatives of trigonometric functions (sin, cos, tan, etc.).
• Inverse Trigonometric Derivatives: Covers derivatives of inverse trigonometric functions (arcsin, arccos, arctan, etc.).

Mastering these basic rules is the crucial first step towards tackling these more advanced techniques. Consistent practice and a thorough understanding of the underlying principles are essential for success in calculus. Remember to break down complex problems into smaller, manageable steps using the rules explained above. This methodical approach will increase your confidence and ability to solve even the most challenging differentiation problems.

This detailed explanation and worked examples serve as a comprehensive substitute for a PDF answer key, providing a deeper understanding of the underlying principles and empowering you to confidently tackle similar problems independently. Remember, consistent practice is the key to mastering differentiation.