2.5 Basic Differentiation Rules Homework Answers
arrobajuarez
Nov 02, 2025 · 10 min read
Table of Contents
Diving into calculus can feel like navigating a maze, especially when faced with differentiation. But fear not! Mastering basic differentiation rules is like equipping yourself with a powerful toolkit, allowing you to solve a multitude of problems. Let's break down the fundamental differentiation rules and tackle some common homework questions to solidify your understanding.
The Power of Differentiation: A Quick Recap
At its core, differentiation is about finding the instantaneous rate of change of a function. Imagine driving a car; differentiation helps us determine your exact speed at any given moment, not just your average speed over a journey. Mathematically, it gives us the slope of the tangent line to a curve at a specific point. This has vast applications across various fields, from physics and engineering to economics and computer science.
Unveiling the 2.5 Basic Differentiation Rules
While there are many differentiation rules, a core set forms the foundation. These are the rules we'll focus on:
- The Constant Rule: The derivative of a constant is always zero.
- The Power Rule: A staple for differentiating polynomial terms.
- The Constant Multiple Rule: Allows you to pull constants out of the differentiation process.
- The Sum and Difference Rule: Simplifies differentiating sums and differences of functions.
- The Product Rule: Essential for differentiating the product of two functions.
- The Quotient Rule: Used to find the derivative of a function divided by another function.
Let's explore each of these in detail with examples.
1. The Constant Rule: When Nothing Changes
The constant rule is perhaps the simplest of all. It states that if f(x) = c, where c is a constant, then f'(x) = 0.
In simpler terms: The derivative of any constant number is always zero. This makes sense because a constant function doesn't change; its rate of change is zero.
Example:
- If f(x) = 7, then f'(x) = 0.
- If y = -3.14, then dy/dx = 0.
- If g(x) = √2, then g'(x) = 0.
2. The Power Rule: Unleashing Exponents
The power rule is a workhorse for differentiating polynomial expressions. It states that if f(x) = x<sup>n</sup>, where n is any real number, then f'(x) = nx<sup>n-1</sup>.
In simpler terms: Multiply the function by the exponent and then subtract 1 from the exponent.
Example 1:
- If f(x) = x<sup>3</sup>, then f'(x) = 3x<sup>2</sup>. (Multiply by 3, reduce the exponent by 1)
Example 2:
- If y = x<sup>-2</sup>, then dy/dx = -2x<sup>-3</sup> which can also be written as dy/dx = -2/x<sup>3</sup>. (Remember that negative exponents mean the reciprocal)
Example 3:
- If g(x) = √x, then first rewrite it as g(x) = x<sup>1/2</sup>. Then, g'(x) = (1/2)x<sup>-1/2</sup> which can also be written as g'(x) = 1/(2√x). (Fractional exponents represent roots)
3. The Constant Multiple Rule: Pulling Out the Numbers
The constant multiple rule simplifies differentiation when a function is multiplied by a constant. It states that if f(x) = c * g(x), where c is a constant, then f'(x) = c * g'(x).
In simpler terms: You can pull the constant out of the derivative and multiply it by the derivative of the remaining function.
Example 1:
- If f(x) = 5x<sup>2</sup>, then f'(x) = 5 * (2x) = 10x.
Example 2:
- If y = -3x<sup>4</sup>, then dy/dx = -3 * (4x<sup>3</sup>) = -12x<sup>3</sup>.
4. The Sum and Difference Rule: Differentiating Piece by Piece
The sum and difference rule allows you to differentiate functions term by term. It states that if f(x) = u(x) + v(x), then f'(x) = u'(x) + v'(x). Similarly, if f(x) = u(x) - v(x), then f'(x) = u'(x) - v'(x).
In simpler terms: The derivative of a sum (or difference) is the sum (or difference) of the derivatives.
Example 1:
- If f(x) = x<sup>3</sup> + 2x<sup>2</sup> - 5x + 1, then f'(x) = 3x<sup>2</sup> + 4x - 5 + 0 = 3x<sup>2</sup> + 4x - 5.
Example 2:
- If y = 4√x - 6x + 7, then first rewrite it as y = 4x<sup>1/2</sup> - 6x + 7. Then, dy/dx = 4 * (1/2)x<sup>-1/2</sup> - 6 + 0 = 2x<sup>-1/2</sup> - 6 = 2/√x - 6.
5. The Product Rule: Handling Multiplication
The product rule is used when differentiating the product of two functions. It states that if f(x) = u(x) * v(x), then f'(x) = u'(x)v(x) + u(x)v'(x).
In simpler terms: The derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. Remember the mnemonic: "First d second plus second d first."
Example:
- Let f(x) = (x<sup>2</sup> + 1)(x<sup>3</sup> - 2x).
- Let u(x) = x<sup>2</sup> + 1, then u'(x) = 2x.
- Let v(x) = x<sup>3</sup> - 2x, then v'(x) = 3x<sup>2</sup> - 2.
- Therefore, f'(x) = (2x)(x<sup>3</sup> - 2x) + (x<sup>2</sup> + 1)(3x<sup>2</sup> - 2).
- Simplifying, f'(x) = 2x<sup>4</sup> - 4x<sup>2</sup> + 3x<sup>4</sup> - 2x<sup>2</sup> + 3x<sup>2</sup> - 2 = 5x<sup>4</sup> - 3x<sup>2</sup> - 2.
6. The Quotient Rule: Dealing with Division
The quotient rule is used when differentiating a function divided by another function. It states that if f(x) = u(x) / v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]<sup>2</sup>.
In simpler terms: The derivative of the quotient is (derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator) all divided by the denominator squared. Remember the mnemonic: "Low d high minus high d low, over low squared."
Example:
- Let f(x) = (x<sup>2</sup>) / (x + 1).
- Let u(x) = x<sup>2</sup>, then u'(x) = 2x.
- Let v(x) = x + 1, then v'(x) = 1.
- Therefore, f'(x) = [(2x)(x + 1) - (x<sup>2</sup>)(1)] / (x + 1)<sup>2</sup>.
- Simplifying, f'(x) = (2x<sup>2</sup> + 2x - x<sup>2</sup>) / (x + 1)<sup>2</sup> = (x<sup>2</sup> + 2x) / (x + 1)<sup>2</sup>.
Tackling Common Homework Problems: Worked Examples
Now, let's put these rules into practice with some typical homework problems.
Problem 1: Find the derivative of f(x) = 3x<sup>5</sup> - 2x<sup>3</sup> + 7x - 4.
Solution:
- Apply the power rule, constant multiple rule, and sum/difference rule.
- f'(x) = 3 * (5x<sup>4</sup>) - 2 * (3x<sup>2</sup>) + 7 * (1) - 0
- f'(x) = 15x<sup>4</sup> - 6x<sup>2</sup> + 7
Problem 2: Find the derivative of y = (x<sup>2</sup> + 3)(2x - 1).
Solution:
- Apply the product rule.
- Let u(x) = x<sup>2</sup> + 3, then u'(x) = 2x.
- Let v(x) = 2x - 1, then v'(x) = 2.
- dy/dx = (2x)(2x - 1) + (x<sup>2</sup> + 3)(2)
- dy/dx = 4x<sup>2</sup> - 2x + 2x<sup>2</sup> + 6
- dy/dx = 6x<sup>2</sup> - 2x + 6
Problem 3: Find the derivative of g(x) = (x + 2) / (x - 1).
Solution:
- Apply the quotient rule.
- Let u(x) = x + 2, then u'(x) = 1.
- Let v(x) = x - 1, then v'(x) = 1.
- g'(x) = [(1)(x - 1) - (x + 2)(1)] / (x - 1)<sup>2</sup>
- g'(x) = (x - 1 - x - 2) / (x - 1)<sup>2</sup>
- g'(x) = -3 / (x - 1)<sup>2</sup>
Problem 4: Find the derivative of f(x) = √x * (x<sup>2</sup> + 1).
Solution:
- First rewrite √x as x<sup>1/2</sup>.
- Apply the product rule.
- Let u(x) = x<sup>1/2</sup>, then u'(x) = (1/2)x<sup>-1/2</sup>.
- Let v(x) = x<sup>2</sup> + 1, then v'(x) = 2x.
- *f'(x) = + *
- f'(x) = (1/2)x<sup>3/2</sup> + (1/2)x<sup>-1/2</sup> + 2x<sup>3/2</sup>
- f'(x) = (5/2)x<sup>3/2</sup> + (1/2)x<sup>-1/2</sup>
- f'(x) = (5/2)x√x + 1/(2√x)
Problem 5: Find the equation of the tangent line to the curve y = x<sup>3</sup> - 2x<sup>2</sup> + x at the point (2, 2).
Solution:
-
Find the derivative:
- dy/dx = 3x<sup>2</sup> - 4x + 1
-
Find the slope of the tangent line at x = 2:
- dy/dx |<sub>x=2</sub> = 3(2)<sup>2</sup> - 4(2) + 1 = 12 - 8 + 1 = 5
- So, the slope m = 5.
-
Use the point-slope form of a line:
- y - y<sub>1</sub> = m(x - x<sub>1</sub>), where (x<sub>1</sub>, y<sub>1</sub>) = (2, 2).
- y - 2 = 5(x - 2)
-
Simplify to slope-intercept form (optional):
- y - 2 = 5x - 10
- y = 5x - 8
Therefore, the equation of the tangent line is y = 5x - 8.
Advanced Applications and Considerations
While these basic rules cover a large range of functions, calculus has far more to offer. Here are a few things to keep in mind as you progress:
- Chain Rule: This is crucial for differentiating composite functions (functions within functions), like sin(x<sup>2</sup>).
- Implicit Differentiation: Used when you can't easily solve for y in terms of x, such as in the equation x<sup>2</sup> + y<sup>2</sup> = 25.
- Higher-Order Derivatives: These are derivatives of derivatives (e.g., the second derivative, third derivative), and they provide information about the concavity and rate of change of the rate of change of a function.
- Applications in Optimization: Derivatives are used to find maximum and minimum values of functions, which is essential in optimization problems (e.g., maximizing profit, minimizing cost).
- Derivatives of Trigonometric, Exponential, and Logarithmic Functions: These functions have their own specific differentiation rules that you'll need to learn.
- L'Hôpital's Rule: A powerful tool for evaluating limits of indeterminate forms using derivatives.
Common Mistakes to Avoid
- Forgetting the Constant Multiple: Make sure to multiply the constant by the entire derivative of the function.
- Misapplying the Power Rule: The power rule only applies to variables raised to a power, not to exponential functions (e.g., 2<sup>x</sup> requires a different rule).
- Incorrectly Applying the Product or Quotient Rule: Pay close attention to the order of operations and the signs in these rules. Writing out u, v, u', and v' separately can help prevent errors.
- Simplifying Too Early: Sometimes, simplifying an expression before differentiating can make the process more complicated.
- Forgetting the "+ C" in Indefinite Integrals: While this article focuses on differentiation, remember that when you find an indefinite integral (the opposite of a derivative), you need to add the constant of integration, C.
Resources for Further Learning
- Khan Academy: Offers free video lessons and practice exercises on calculus and differentiation.
- MIT OpenCourseware: Provides access to lecture notes, problem sets, and exams from MIT calculus courses.
- Paul's Online Math Notes: A comprehensive website with clear explanations and examples of calculus concepts.
- Your Textbook and Professor: Don't forget the resources available through your own course!
Conclusion: Mastering the Basics is Key
Understanding and mastering these basic differentiation rules is essential for success in calculus and related fields. Practice is key! The more you work through problems, the more comfortable you'll become with applying these rules and recognizing when to use them. Remember to break down complex problems into smaller, manageable steps, and don't be afraid to seek help when you need it. With dedication and practice, you'll be able to confidently tackle any differentiation challenge that comes your way. Happy differentiating!
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