Calculating Partial Derivatives: A Step-by-Step Guide

Hey everyone! Today, we're diving into the world of partial derivatives and tackling a cool problem. We're given a function, and our mission is to calculate a specific partial derivative at a particular point. Don't worry if it sounds intimidating; we'll break it down step by step and make it super understandable. So, grab your coffee, and let's get started!

Understanding the Function and Partial Derivatives

First things first, let's get acquainted with our function: $f(x, y)=\begin{cases}\frac{x^3 y-x y^3}{x^2+y^2} & \text { if } & (x, y) \neq(0,0) \\0 & \text { if } & (x, y)=(0,0)\end{cases}$. This function, f(x, y), is defined differently depending on whether the point (x, y) is the origin (0, 0) or not. This is a common situation in multivariable calculus, and it's important to keep this in mind. Now, the main star of our show is the partial derivative. In essence, a partial derivative tells us how a function changes with respect to one variable while holding the other variable(s) constant. Think of it like this: If you're walking along a hilly landscape (your function), the partial derivative is like asking, "How steep is the hill if I only move east/west?" or "How steep is it if I only move north/south?" We denote the partial derivative of f with respect to x as $\frac{\partial f}{\partial x}$, and it represents the rate of change of f as x changes, while y remains fixed. Similarly, $\frac{\partial f}{\partial y}$ represents the rate of change of f as y changes, while x remains fixed. To find these partial derivatives, we'll use the limit definition or, in most cases, apply differentiation rules we learned in single-variable calculus, but with a twist. We treat the other variable(s) as constants during differentiation. Understanding these core concepts is crucial because it forms the foundation for tackling more complex problems and applications in various fields, including physics, engineering, economics, and data science. So, when dealing with partial derivatives, remember the core idea: We're investigating the function's rate of change with respect to one variable at a time, keeping everything else constant. This isolation allows us to understand how each variable influences the function's behavior.

The Importance of the Point (2,0)

Now, about that point (2, 0). We're asked to evaluate $\frac{\partial f}{\partial x}$ specifically at the point (2, 0), written as $\left.\frac{\partial f}{\partial x}\right|_{(2,0)}$. This is where things get interesting! The point (2, 0) gives us the exact coordinates where we want to know the rate of change of f with respect to x. Because (2, 0) is not (0, 0), we will use the first part of the function definition to get the partial derivative. This means we'll differentiate the expression $\frac{x^3 y-x y^3}{x^2+y^2}$ with respect to x, and then we'll plug in x = 2 and y = 0 into the derivative. In essence, the partial derivative at a specific point is a numerical value that tells us the instantaneous rate of change of the function at that precise location. This can give us valuable information, such as the slope of a tangent plane or the direction in which the function is increasing or decreasing the most. For instance, in real-world scenarios, this might correspond to the rate of change of temperature across a surface, the rate of change of profit based on product sales, or even the gradient of a neural network in machine learning. Hence, calculating partial derivatives at specific points is critical to understanding the function's behavior and applying these concepts to practical problems. Remember, the point (2, 0) provides the specific context, directing us to assess the function's behavior at that particular location.

Calculating the Partial Derivative

Alright, let's roll up our sleeves and get down to business! Since we're not at the origin, we'll use the definition of f(x, y) when (x, y) ≠ (0, 0). This means we are working with: $f(x, y) = \frac{x^3y - xy^3}{x^2 + y^2}$. Our goal is to find $\frac{\partial f}{\partial x}$. To do this, we'll treat y as a constant and differentiate f with respect to x. This involves applying the quotient rule, which states that the derivative of $\frac{u}{v}$ is $\frac{v(\frac{\partial u}{\partial x}) - u(\frac{\partial v}{\partial x})}{v^2}$, where u and v are functions of x and y. So, let's identify u and v: u = x³y - xy³ and v = x² + y². Now, we differentiate u and v with respect to x: $\frac{\partial u}{\partial x} = 3x^2y - y^3$ and $\frac{\partial v}{\partial x} = 2x$. Using the quotient rule, we get:

$\frac{\partial f}{\partial x} = \frac{(x^2 + y^2)(3x^2y - y^3) - (x^3y - xy^3)(2x)}{(x^2 + y^2)^2}$.

This might look a bit messy, but trust me, it's manageable! The next step is to simplify this expression. Expand the numerator and combine like terms to make it easier to work with. Remember, we want to find the partial derivative at the point (2, 0), so we will plug in those values for x and y after simplifying the derivative. Simplification reduces the chances of making arithmetic errors when substituting the values of x and y. This is where your algebra skills come into play. Take your time, be methodical, and double-check your work to ensure accuracy. The simplification process may involve several steps, but it's essential to arrive at a manageable expression. After simplifying, we can substitute the values of x and y to obtain the final numerical result. Keeping track of signs and exponents is also important. So, while the initial expression might seem complicated, simplifying it systematically is key to reaching the solution. This process not only makes the calculation easier but also gives you a deeper understanding of the function's behavior, allowing you to appreciate the elegant math behind it all.

Simplifying the Expression

Let's continue with the simplification of the partial derivative. Recall that we have: $\frac{\partial f}{\partial x} = \frac{(x^2 + y^2)(3x^2y - y^3) - (x^3y - xy^3)(2x)}{(x^2 + y^2)^2}$. Expanding the numerator, we get:

$\frac{\partial f}{\partial x} = \frac{3x^4y - x^2y^3 + 3x^2y^3 - y^5 - 2x^4y + 2x^2y^3}{(x^2 + y^2)^2}$.

Now, let's combine the like terms in the numerator:

$\frac{\partial f}{\partial x} = \frac{x^4y + 4x^2y^3 - y^5}{(x^2 + y^2)^2}$.

Looks much cleaner, right? This simplified form is what we'll use to evaluate the partial derivative at the point (2, 0). This simplification step ensures that when we substitute the values of x and y, our calculations are as straightforward as possible. It is also good practice, because it helps reduce the chances of making arithmetic errors. We are essentially reorganizing the terms to create an expression that is ready for calculation at any specific point. The effort spent in simplifying the expression is well worth it, because it greatly increases the speed and accuracy of the final calculation. Furthermore, a simplified expression offers a clearer view of the function's behavior. We can see how the different powers of x and y interact to determine the rate of change of the function in the x-direction. So, by carefully simplifying the expression, we've prepared ourselves to complete the final calculation and find the specific value of the partial derivative at our point of interest.

Evaluating at (2, 0)

Now comes the exciting part! We've found the general expression for $\frac{\partial f}{\partial x}$, and now we're going to plug in the values x = 2 and y = 0 to find $\left.\frac{\partial f}{\partial x}\right|_{(2,0)}$. Recall our simplified expression: $\frac{\partial f}{\partial x} = \frac{x^4y + 4x^2y^3 - y^5}{(x^2 + y^2)^2}$. Let's substitute x = 2 and y = 0:

$\left.\frac{\partial f}{\partial x}\right|_{(2,0)} = \frac{(2)^4(0) + 4(2)^2(0)^3 - (0)^5}{((2)^2 + (0)^2)^2}$.

This simplifies to:

$\left.\frac{\partial f}{\partial x}\right|_{(2,0)} = \frac{0 + 0 - 0}{(4 + 0)^2} = \frac{0}{16} = 0$.

And there you have it, guys! The partial derivative of f with respect to x at the point (2, 0) is 0. This means that at the point (2, 0), the function f is not changing with respect to x. This can also be interpreted geometrically, meaning that the tangent plane to the surface defined by f at the point (2, 0, f(2, 0)) is parallel to the yz-plane. This insight provides valuable information about the function's behavior at this specific location. We should keep in mind that the value of the partial derivative is location-dependent. If we chose a different point, say (1, 1), we would obtain a different value. So the result we've obtained applies only to the point (2, 0). Knowing the rate of change is 0 tells us that if you slightly move in the x-direction from this point, the value of the function won't change at all. This is useful information when analyzing the behavior of the function, because we know the function is neither increasing nor decreasing with respect to x at this point. This complete process shows how partial derivatives help us understand the behavior of multivariable functions, making the problem easier to solve. Congratulations, we've successfully calculated the partial derivative!

Conclusion

We did it, everyone! We successfully calculated the partial derivative of the function with respect to x at the point (2, 0). We started by understanding the function and the concept of partial derivatives. Then, we applied the quotient rule, simplified the expression, and finally, plugged in the specific values to get our answer: 0. This entire exercise helps you build a solid foundation in multivariable calculus. Keep practicing, and you'll become a pro in no time! Remember, these concepts are widely applicable, and understanding them will serve you well in various fields, from science to engineering. Keep exploring and happy calculating!