2.5 Basic Differentiation Rules Homework Answers

2.5 Basic Differentiation Rules: Homework Answers and Deeper Understanding

Calculus, particularly differentiation, can feel daunting at first. However, mastering the basic rules is the key to unlocking more complex concepts. This article provides comprehensive answers and explanations for common homework problems related to the 2.5 basic differentiation rules, going beyond simple solutions to build a strong intuitive understanding. We'll cover the power rule, constant multiple rule, sum/difference rule, and delve into practical applications.

1. The Power Rule: The Foundation of Differentiation

The power rule is the cornerstone of differentiation. It states that the derivative of xⁿ is nx^n-1, where 'n' is any real number. Let's tackle some examples:

Problem 1: Find the derivative of f(x) = x⁵.

Solution: Applying the power rule, we get f'(x) = 5x^5-1 = 5x⁴. Simple, right? The exponent becomes the coefficient, and the new exponent is one less than the original.

Problem 2: Find the derivative of g(x) = x^-3.

Solution: The power rule works equally well for negative exponents. g'(x) = -3x^-3-1 = -3x^-4. Notice how the negative sign carries through.

Problem 3: Find the derivative of h(x) = √x.

Solution: Remember that √x is the same as x^1/2. Applying the power rule: h'(x) = (1/2)x^(1/2)-1 = (1/2)x^-1/2 = 1/(2√x). This highlights how the power rule handles fractional exponents.

Problem 4 (Challenge): Find the derivative of k(x) = x^π.

Solution: The power rule is not limited to integer exponents! k'(x) = πx^π-1. This emphasizes the generality and power of the rule.

2. The Constant Multiple Rule: Scaling Derivatives

The constant multiple rule states that the derivative of a constant times a function is the constant times the derivative of the function. In simpler terms: d/dx[cf(x)] = c * f'(x), where 'c' is a constant.

Problem 5: Find the derivative of f(x) = 3x⁴.

Solution: Using the constant multiple rule and the power rule: f'(x) = 3 * (4x³) = 12x³. We differentiate the x⁴ part first, then multiply by the constant 3.

Problem 6: Find the derivative of g(x) = -2/x².

Solution: Rewrite g(x) as -2x^-2. Then, using the constant multiple and power rules: g'(x) = -2 * (-2x^-3) = 4x^-3 = 4/x³.

Problem 7: Find the derivative of h(x) = πx⁷.

Solution: h'(x) = π * (7x⁶) = 7πx⁶. Constants, including π, are treated as they are.

3. The Sum and Difference Rules: Handling Multiple Terms

The sum rule states that the derivative of a sum of functions is the sum of their derivatives. Similarly, the difference rule states that the derivative of a difference of functions is the difference of their derivatives. In short: d/dx[f(x) ± g(x)] = f'(x) ± g'(x).

Problem 8: Find the derivative of f(x) = x³ + 2x² - 5x + 7.

Solution: Applying the sum/difference rule and power rule to each term: f'(x) = 3x² + 4x - 5. The derivative of the constant 7 is 0.

Problem 9: Find the derivative of g(x) = 4x⁵ - √x + 1/x.

Solution: Rewrite g(x) as 4x⁵ - x^1/2 + x^-1. Then: g'(x) = 20x⁴ - (1/2)x^-1/2 - x^-2 = 20x⁴ - 1/(2√x) - 1/x². We handled each term separately.

Problem 10 (Challenge): Find the derivative of h(x) = (x² + 3x - 1)(2x + 5).

Solution: Before applying differentiation rules, expand the expression: h(x) = 2x³ + 11x² + 13x -5. Now differentiate term by term: h'(x) = 6x² + 22x + 13. This shows that sometimes algebraic manipulation is needed before applying differentiation rules.

4. Combining the Rules: Tackling More Complex Functions

Real-world applications often involve functions that require combining multiple differentiation rules.

Problem 11: Find the derivative of f(x) = 2x³(x² - 4x + 1).

Solution: First expand: f(x) = 2x⁵ - 8x⁴ + 2x³. Then differentiate term by term: f'(x) = 10x⁴ - 32x³ + 6x².

Problem 12: Find the derivative of g(x) = (3x² + 2x - 1)/(x).

Solution: Rewrite as separate terms: g(x) = 3x + 2 - 1/x = 3x + 2 - x^-1. Then differentiate: g'(x) = 3 + x^-2 = 3 + 1/x².

Problem 13 (Challenge): Find the derivative of h(x) = [5x³ + (2/x) ]².

Solution: This problem requires the chain rule (which we haven't explicitly covered, but is implied in the question's context of 2.5 basic rules). While we avoid detailed explanation of the chain rule here, understanding the underlying principles from this question is beneficial for future learning. The solution requires first expanding the square. This leads to: h(x) = 25x⁶ + 20x² + 4x^-2. Now, differentiation is straightforward: h'(x) = 150x⁵ + 40x - 8x^-3 = 150x⁵ + 40x - 8/x³.

5. Applications and Further Exploration

The 2.5 basic differentiation rules are not merely abstract mathematical concepts. They form the basis for numerous applications in various fields:

• Physics: Calculating velocity and acceleration from displacement functions.
• Engineering: Optimizing designs and finding maximum/minimum values.
• Economics: Determining marginal cost, revenue, and profit.
• Computer Science: Developing algorithms for optimization and machine learning.

Beyond these basic rules, more advanced techniques like the chain rule, product rule, and quotient rule will allow you to differentiate even more complex functions. These rules build directly upon the foundation you've laid here.

Conclusion:

Mastering the 2.5 basic differentiation rules is crucial for success in calculus and its applications. This article provides not only answers to common homework problems but also a deeper understanding of the underlying concepts. By practicing these rules and exploring their applications, you'll build a solid foundation for tackling more challenging problems in the future. Remember that consistent practice and a clear understanding of the underlying principles are key to success in calculus. Don't hesitate to work through additional problems to solidify your grasp of these fundamental rules. The more you practice, the more intuitive differentiation will become.