Partial Derivatives Worksheet With Answers

Partial Derivatives Worksheet: A Comprehensive Guide with Solved Examples

Understanding partial derivatives is crucial for anyone studying calculus, particularly in multivariable calculus. This worksheet provides a comprehensive guide, covering various aspects of partial derivatives, from basic computations to more advanced applications. It's designed to help you master the concept and build a solid foundation for further study in fields like physics, engineering, and economics. This worksheet includes a range of problems with detailed solutions, allowing you to check your understanding and identify areas needing further practice.

Introduction to Partial Derivatives

Partial derivatives extend the concept of ordinary derivatives to functions of multiple variables. While an ordinary derivative measures the instantaneous rate of change of a function with respect to a single variable, a partial derivative measures the instantaneous rate of change of a multivariable function with respect to one variable, holding all other variables constant. This "holding constant" is the key difference and the source of many initial difficulties.

Consider a function of two variables, f(x, y). The partial derivative with respect to x, denoted as ∂f/∂x or f_x, is found by treating y as a constant and differentiating f(x, y) with respect to x using the standard rules of differentiation. Similarly, the partial derivative with respect to y, denoted as ∂f/∂y or f_y, is found by treating x as a constant and differentiating f(x, y) with respect to y.

Steps to Calculate Partial Derivatives

Calculating partial derivatives follows a systematic approach:

1. Identify the variable of differentiation: Determine which variable you are differentiating with respect to (e.g., x or y).

2. Treat other variables as constants: This is the crucial step. Consider all other variables as fixed numerical values, not as variables themselves.

3. Apply standard differentiation rules: Use the usual rules of differentiation (power rule, product rule, quotient rule, chain rule, etc.) to differentiate the function with respect to the chosen variable, treating the other variables as constants.

4. Simplify the result: Simplify the resulting expression as much as possible.

Worked Examples: Basic Partial Derivatives

Let's illustrate the process with some examples:

Example 1:

Find the partial derivatives ∂f/∂x and ∂f/∂y for the function f(x, y) = x²y + 3xy² + 2x – y.

Solution:

• ∂f/∂x: Treat y as a constant. The derivative of x²y with respect to x is 2xy. The derivative of 3xy² with respect to x is 3y². The derivative of 2x with respect to x is 2. The derivative of -y with respect to x is 0. Therefore, ∂f/∂x = 2xy + 3y² + 2.

• ∂f/∂y: Treat x as a constant. The derivative of x²y with respect to y is x². The derivative of 3xy² with respect to y is 6xy. The derivative of 2x with respect to y is 0. The derivative of -y with respect to y is -1. Therefore, ∂f/∂y = x² + 6xy – 1.

Example 2:

Find the partial derivatives ∂z/∂x and ∂z/∂y for the function z = e^(x²y) + ln(xy).

Solution:

• ∂z/∂x: Treat y as a constant. Using the chain rule, the derivative of e^(x²y) with respect to x is (2xy)e^(x²y). The derivative of ln(xy) with respect to x is (1/xy) * y = 1/x. Therefore, ∂z/∂x = 2xye^(x²y) + 1/x.

• ∂z/∂y: Treat x as a constant. Using the chain rule, the derivative of e^(x²y) with respect to y is (x²)e^(x²y). The derivative of ln(xy) with respect to y is (1/xy) * x = 1/y. Therefore, ∂z/∂y = x²e^(x²y) + 1/y.

Example 3:

Find the partial derivatives ∂f/∂x and ∂f/∂y for the function f(x, y) = (x² + y²) / (x – y).

Solution:

• ∂f/∂x: Using the quotient rule (remembering that y is constant), we get: [(2x)(x-y) - (x²+y²)(1)] / (x-y)² = (2x² - 2xy - x² - y²) / (x-y)² = (x² - 2xy - y²) / (x-y)²

• ∂f/∂y: Using the quotient rule (remembering that x is constant), we get: [(2y)(x-y) - (x²+y²)(-1)] / (x-y)² = (2xy - 2y² + x² + y²) / (x-y)² = (x² + 2xy - y²) / (x-y)²

Worked Examples: More Advanced Partial Derivatives

These examples introduce more complex scenarios involving multiple variables and advanced differentiation techniques.

Example 4 (Chain Rule):

Let z = x² + y², where x = u + v and y = u – v. Find ∂z/∂u and ∂z/∂v.

Solution:

We can use the chain rule. First, express z directly in terms of u and v:

z = (u+v)² + (u-v)² = u² + 2uv + v² + u² - 2uv + v² = 2u² + 2v²

Now we can find the partial derivatives:

• ∂z/∂u = 4u
• ∂z/∂v = 4v

Example 5 (Implicit Differentiation):

Find ∂z/∂x and ∂z/∂y given the equation x² + y² + z² = 1.

Solution:

We use implicit differentiation. Remember to treat the variables we're not differentiating with respect to as constants.

• ∂/∂x: 2x + 2z(∂z/∂x) = 0 => ∂z/∂x = -x/z

• ∂/∂y: 2y + 2z(∂z/∂y) = 0 => ∂z/∂y = -y/z

Example 6 (Higher-Order Partial Derivatives):

Find the second-order partial derivatives (f_xx, f_yy, f_xy, f_yx) for f(x, y) = x³y² + 2x²y + x + y.

Solution:

First, find the first-order partial derivatives:

• f_x = 3x²y² + 4xy + 1
• f_y = 2x³y + 2x²

Then, find the second-order partial derivatives:

• f_xx = 6xy² + 4y
• f_yy = 2x³
• f_xy = 6x²y + 4x
• f_yx = 6x²y + 4x

Note that in this example, f_xy = f_yx. This is true for most "well-behaved" functions; Clairaut's theorem states this equality under certain continuity conditions.

Frequently Asked Questions (FAQ)

• Q: What is the difference between a partial derivative and an ordinary derivative?

  • A: An ordinary derivative measures the rate of change of a function with respect to a single variable. A partial derivative measures the rate of change of a multivariable function with respect to one variable, holding all other variables constant.

• Q: Why do we treat other variables as constants when calculating partial derivatives?

  • A: We treat other variables as constants to isolate the effect of the variable we are differentiating with respect to. This allows us to focus on the instantaneous rate of change in one direction only.

• Q: What are higher-order partial derivatives?

  • A: Higher-order partial derivatives are found by taking partial derivatives of partial derivatives. For instance, taking the partial derivative of ∂f/∂x with respect to x gives ∂²f/∂x².

• Q: What are some applications of partial derivatives?

  • A: Partial derivatives have vast applications in various fields, including:
    • Physics: Calculating rates of change in systems with multiple variables (e.g., thermodynamics, electromagnetism).
    • Engineering: Optimizing designs, analyzing stress and strain in materials.
    • Economics: Analyzing marginal productivity, optimizing resource allocation.
    • Machine Learning: Gradient descent algorithms for optimizing models.

Conclusion

This worksheet provides a solid foundation for understanding partial derivatives. Remember the key concept: treat all variables except the one you're differentiating with respect to as constants. Consistent practice with a variety of examples is crucial to mastering this fundamental concept in multivariable calculus. By working through these examples and applying the steps outlined, you will gain confidence and proficiency in calculating and interpreting partial derivatives. Continue practicing to solidify your understanding and prepare for more advanced topics in calculus and related fields. Remember to consult your textbook and lecture notes for additional examples and explanations. Good luck!