Find The First Partial Derivatives Of The Function
arrobajuarez
Nov 30, 2025 · 10 min read
Table of Contents
Let's explore the fascinating world of partial derivatives! Understanding how to find the first partial derivatives of a function is a crucial skill in multivariable calculus, with applications across various fields like physics, engineering, economics, and computer graphics. This article will guide you through the concept, providing a step-by-step approach with examples to solidify your understanding.
Introduction to Partial Derivatives
In single-variable calculus, we deal with functions of the form y = f(x), where y depends on a single independent variable x. The derivative, dy/dx, tells us the rate of change of y with respect to x. However, in many real-world scenarios, quantities depend on multiple variables. For instance, the temperature at a point on a metal plate depends on both the x and y coordinates of the point.
This is where partial derivatives come in. A partial derivative measures the rate of change of a multivariable function with respect to one of its variables, while holding the other variables constant. This allows us to analyze the influence of each variable independently.
Notation
The notation for partial derivatives can vary, but the most common is:
- For a function f(x, y):
- The partial derivative of f with respect to x is denoted as ∂f/∂x, f<sub>x</sub>, or D<sub>x</sub>f.
- The partial derivative of f with respect to y is denoted as ∂f/∂y, f<sub>y</sub>, or D<sub>y</sub>f.
Similarly, for a function f(x, y, z):
- The partial derivative of f with respect to x is denoted as ∂f/∂x, f<sub>x</sub>, or D<sub>x</sub>f.
- The partial derivative of f with respect to y is denoted as ∂f/∂y, f<sub>y</sub>, or D<sub>y</sub>f.
- The partial derivative of f with respect to z is denoted as ∂f/∂z, f<sub>z</sub>, or D<sub>z</sub>f.
The symbol ∂ (a rounded d) is used to distinguish partial derivatives from ordinary derivatives.
Step-by-Step Guide to Finding First Partial Derivatives
Here's a detailed guide on how to calculate first partial derivatives:
1. Identify the Function and its Variables:
- Clearly define the function you're working with, such as f(x, y) = x<sup>2</sup> + xy + y<sup>2</sup>.
- Identify all the independent variables in the function (e.g., x and y in the example above).
2. Choose the Variable for Differentiation:
- Decide which variable you want to differentiate with respect to first (e.g., x or y).
3. Treat Other Variables as Constants:
- This is the crucial step! When differentiating with respect to one variable, treat all other variables as if they were constant numbers. For example, if differentiating f(x, y) with respect to x, treat y as a constant.
4. Apply Standard Differentiation Rules:
- Use the familiar rules of differentiation from single-variable calculus, such as:
- Power Rule: d/dx (x<sup>n</sup>) = nx<sup>n-1</sup>
- Constant Multiple Rule: d/dx (cf(x)) = c * d/dx (f(x))
- Sum/Difference Rule: d/dx (f(x) ± g(x)) = d/dx (f(x)) ± d/dx (g(x))
- Product Rule: d/dx (f(x)g(x)) = f'(x)g(x) + f(x)g'(x)
- Quotient Rule: d/dx (f(x)/g(x)) = (g(x)f'(x) - f(x)g'(x)) / (g(x))<sup>2</sup>
- Chain Rule: d/dx (f(g(x))) = f'(g(x)) * g'(x)
- Derivatives of Trigonometric Functions: d/dx (sin(x)) = cos(x), d/dx (cos(x)) = -sin(x), etc.
- Derivatives of Exponential and Logarithmic Functions: d/dx (e<sup>x</sup>) = e<sup>x</sup>, d/dx (ln(x)) = 1/x
5. Simplify the Result:
- After applying the differentiation rules, simplify the expression to obtain the partial derivative.
6. Repeat for Other Variables:
- Repeat steps 2-5 for each of the other independent variables in the function.
Examples with Detailed Solutions
Let's illustrate the process with several examples:
Example 1: f(x, y) = x<sup>2</sup> + xy + y<sup>2</sup>
-
Find ∂f/∂x:
- Treat y as a constant.
- ∂f/∂x = ∂/∂x (x<sup>2</sup>) + ∂/∂x (xy) + ∂/∂x (y<sup>2</sup>)
- ∂f/∂x = 2x + y + 0 (Since y<sup>2</sup> is a constant with respect to x, its derivative is 0)
- ∂f/∂x = 2x + y
-
Find ∂f/∂y:
- Treat x as a constant.
- ∂f/∂y = ∂/∂y (x<sup>2</sup>) + ∂/∂y (xy) + ∂/∂y (y<sup>2</sup>)
- ∂f/∂y = 0 + x + 2y (Since x<sup>2</sup> is a constant with respect to y, its derivative is 0)
- ∂f/∂y = x + 2y
Example 2: f(x, y) = sin(x)cos(y)
-
Find ∂f/∂x:
- Treat y as a constant. Therefore, cos(y) is also a constant.
- ∂f/∂x = ∂/∂x (sin(x)cos(y)) = cos(y) * ∂/∂x (sin(x))
- ∂f/∂x = cos(y) * cos(x) = cos(x)cos(y)
-
Find ∂f/∂y:
- Treat x as a constant. Therefore, sin(x) is also a constant.
- ∂f/∂y = ∂/∂y (sin(x)cos(y)) = sin(x) * ∂/∂y (cos(y))
- ∂f/∂y = sin(x) * (-sin(y)) = -sin(x)sin(y)
Example 3: f(x, y) = e<sup>(x<sup>2</sup>y)</sup>
-
Find ∂f/∂x:
- Treat y as a constant. We need to use the chain rule.
- ∂f/∂x = ∂/∂x (e<sup>(x<sup>2</sup>y)</sup>) = e<sup>(x<sup>2</sup>y)</sup> * ∂/∂x (x<sup>2</sup>y)
- ∂f/∂x = e<sup>(x<sup>2</sup>y)</sup> * (2xy) = 2xye<sup>(x<sup>2</sup>y)</sup>
-
Find ∂f/∂y:
- Treat x as a constant. We need to use the chain rule.
- ∂f/∂y = ∂/∂y (e<sup>(x<sup>2</sup>y)</sup>) = e<sup>(x<sup>2</sup>y)</sup> * ∂/∂y (x<sup>2</sup>y)
- ∂f/∂y = e<sup>(x<sup>2</sup>y)</sup> * (x<sup>2</sup>) = x<sup>2</sup>e<sup>(x<sup>2</sup>y)</sup>
Example 4: f(x, y, z) = x<sup>2</sup>yz + xy<sup>3</sup>z<sup>2</sup> + z<sup>4</sup>
-
Find ∂f/∂x:
- Treat y and z as constants.
- ∂f/∂x = ∂/∂x (x<sup>2</sup>yz) + ∂/∂x (xy<sup>3</sup>z<sup>2</sup>) + ∂/∂x (z<sup>4</sup>)
- ∂f/∂x = 2xyz + y<sup>3</sup>z<sup>2</sup> + 0
- ∂f/∂x = 2xyz + y<sup>3</sup>z<sup>2</sup>
-
Find ∂f/∂y:
- Treat x and z as constants.
- ∂f/∂y = ∂/∂y (x<sup>2</sup>yz) + ∂/∂y (xy<sup>3</sup>z<sup>2</sup>) + ∂/∂y (z<sup>4</sup>)
- ∂f/∂y = x<sup>2</sup>z + 3xy<sup>2</sup>z<sup>2</sup> + 0
- ∂f/∂y = x<sup>2</sup>z + 3xy<sup>2</sup>z<sup>2</sup>
-
Find ∂f/∂z:
- Treat x and y as constants.
- ∂f/∂z = ∂/∂z (x<sup>2</sup>yz) + ∂/∂z (xy<sup>3</sup>z<sup>2</sup>) + ∂/∂z (z<sup>4</sup>)
- ∂f/∂z = x<sup>2</sup>y + 2xy<sup>3</sup>z + 4z<sup>3</sup>
Example 5: f(x, y) = ln(x<sup>2</sup> + y<sup>2</sup>)
-
Find ∂f/∂x:
- Treat y as a constant. We'll need the chain rule.
- ∂f/∂x = ∂/∂x (ln(x<sup>2</sup> + y<sup>2</sup>)) = (1 / (x<sup>2</sup> + y<sup>2</sup>)) * ∂/∂x (x<sup>2</sup> + y<sup>2</sup>)
- ∂f/∂x = (1 / (x<sup>2</sup> + y<sup>2</sup>)) * (2x + 0)
- ∂f/∂x = 2x / (x<sup>2</sup> + y<sup>2</sup>)
-
Find ∂f/∂y:
- Treat x as a constant. We'll need the chain rule.
- ∂f/∂y = ∂/∂y (ln(x<sup>2</sup> + y<sup>2</sup>)) = (1 / (x<sup>2</sup> + y<sup>2</sup>)) * ∂/∂y (x<sup>2</sup> + y<sup>2</sup>)
- ∂f/∂y = (1 / (x<sup>2</sup> + y<sup>2</sup>)) * (0 + 2y)
- ∂f/∂y = 2y / (x<sup>2</sup> + y<sup>2</sup>)
Higher-Order Partial Derivatives
Just as with ordinary derivatives, we can take derivatives of partial derivatives, resulting in higher-order partial derivatives. For example, if we have f(x, y):
- f<sub>xx</sub> = ∂<sup>2</sup>f/∂x<sup>2</sup> = ∂/∂x (∂f/∂x) (The second partial derivative with respect to x)
- f<sub>yy</sub> = ∂<sup>2</sup>f/∂y<sup>2</sup> = ∂/∂y (∂f/∂y) (The second partial derivative with respect to y)
- f<sub>xy</sub> = ∂<sup>2</sup>f/∂y∂x = ∂/∂y (∂f/∂x) (The mixed partial derivative - differentiate with respect to x first, then y)
- f<sub>yx</sub> = ∂<sup>2</sup>f/∂x∂y = ∂/∂x (∂f/∂y) (The mixed partial derivative - differentiate with respect to y first, then x)
Clairaut's Theorem (Equality of Mixed Partials):
Under certain conditions (specifically, if the second partial derivatives are continuous), the order of differentiation doesn't matter for mixed partial derivatives. That is:
- f<sub>xy</sub> = f<sub>yx</sub>
This theorem is incredibly useful because it allows us to choose the order of differentiation that's easiest.
Example: f(x, y) = x<sup>3</sup>y<sup>2</sup> + x<sup>4</sup>
- f<sub>x</sub> = 3x<sup>2</sup>y<sup>2</sup> + 4x<sup>3</sup>
- f<sub>y</sub> = 2x<sup>3</sup>y
- f<sub>xy</sub> = ∂/∂y (3x<sup>2</sup>y<sup>2</sup> + 4x<sup>3</sup>) = 6x<sup>2</sup>y
- f<sub>yx</sub> = ∂/∂x (2x<sup>3</sup>y) = 6x<sup>2</sup>y
As you can see, f<sub>xy</sub> = f<sub>yx</sub>, illustrating Clairaut's Theorem.
Applications of Partial Derivatives
Partial derivatives are used extensively in various fields:
- Optimization: Finding maxima and minima of multivariable functions, crucial in economics, engineering design, and machine learning. Critical points occur where all first partial derivatives are zero or undefined.
- Physics: Describing the behavior of systems involving multiple variables, such as heat flow, fluid dynamics, and electromagnetism. For example, the gradient of a scalar field (like temperature) is a vector composed of partial derivatives.
- Engineering: Designing structures, optimizing processes, and analyzing systems with multiple inputs.
- Economics: Modeling economic systems and optimizing resource allocation. For instance, marginal utility is a partial derivative representing the change in utility from consuming one more unit of a good.
- Computer Graphics: Creating realistic images and animations by calculating how light interacts with surfaces. Partial derivatives are used in shading models and texture mapping.
- Machine Learning: Training neural networks by adjusting weights to minimize a loss function. This is done using gradient descent, which relies on partial derivatives.
Common Mistakes to Avoid
- Forgetting to treat other variables as constants: This is the most common mistake. Always remember that when taking a partial derivative with respect to one variable, all other variables are treated as constants.
- Incorrectly applying differentiation rules: Make sure you are comfortable with the standard differentiation rules from single-variable calculus.
- Not using the chain rule when necessary: If the function involves a composite function, remember to apply the chain rule.
- Not simplifying the result: Always simplify the expression after taking the partial derivative.
- Confusing partial derivatives with ordinary derivatives: Use the correct notation (∂ for partial derivatives, d for ordinary derivatives).
Practice Problems
To solidify your understanding, try these practice problems:
- f(x, y) = x<sup>3</sup> - 3xy + y<sup>2</sup>
- f(x, y) = xe<sup>y</sup> + ye<sup>x</sup>
- f(x, y) = tan<sup>-1</sup>(y/x)
- f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> + xyz
- f(x, y) = (x<sup>2</sup> + y<sup>2</sup>)<sup>1/2</sup>
(Try finding all first partial derivatives for each function).
Conclusion
Finding the first partial derivatives of a function is a fundamental concept in multivariable calculus. By following the step-by-step guide and practicing with examples, you can master this skill. Remember to treat other variables as constants, apply the standard differentiation rules, and simplify your results. Understanding partial derivatives opens the door to solving complex problems in various fields, from physics and engineering to economics and computer science. So, keep practicing, and you'll become proficient in this powerful tool!
Latest Posts
Latest Posts
-
Assume That Both Populations Are Normally Distributed
Nov 30, 2025
-
Which Of The Following Glands Synthesizes Antidiuretic Hormone
Nov 30, 2025
-
Label The Features Of Simple Cuboidal Epithelium
Nov 30, 2025
-
Find The First Partial Derivatives Of The Function
Nov 30, 2025
-
A Countershaft Carrying Two V Belt
Nov 30, 2025
Related Post
Thank you for visiting our website which covers about Find The First Partial Derivatives Of The Function . We hope the information provided has been useful to you. Feel free to contact us if you have any questions or need further assistance. See you next time and don't miss to bookmark.