Question

Use the function to prove that (a) \(f_{x}(0,0)\) and \(f_{y}(\mathbf{0}, \mathbf{0})\) exist, and (b) \(f\) is not differentiable at \((\mathbf{0}, \mathbf{0})\). \(f(x, y)=\left\\{\begin{array}{ll}\frac{5 x^{2} y}{x^{3}+y^{3}}, & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0)\end{array}\right.\)

Short answer

Both partial derivatives \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist and are equal to zero. However, the function is not differentiable at point (0,0) because the limit as \(x\) and \(y\) approach to that point is not same along all paths.

Step-by-step solution

Calculate partial derivative with respect to x

We need to calculate the partial derivative of the function with respect to \(x\) and then compute its value at (0,0).\nFirst, calculate the derivative of the function \(f(x, y)\) with respect to \(x\), treat \(y\) as a constant.\n\(f_{x}(x, y)=\frac{5y(x^{2}+3x^{2}y^{2}-y^{4})}{(x^{2}+y^{2})^{3}}\). To find \(f_{x}(0,0)\), take the limit as \(x\) approaches zero while keeping \(y\) constant. \(lim_{x\rightarrow 0} f_{x}(x, y)=0\). This limit exists and is equal to zero, so \(f_{x}(0,0)\) exists. Note that the other part of the function is not defined at (0,0)

Calculate partial derivative with respect to y

Similar to Step 1, we need to calculate the partial derivative of the function with respect to \(y\) and then compute its value at (0,0).\nThe derivative of the function \(f(x, y)\) with respect to \(y\), treating \(x\) as a constant is:\(f_{y}(x, y)=\frac{5x(x^{2}-3x^{2}y^{2}+y^{4})}{(x^{2}+y^{2})^{3}}\).To find \(f_{y}(0,0)\), take the limit as \(y\) approaches zero while keeping \(x\) constant. \(lim_{y\rightarrow 0} f_{y}(x, y)=0\). This limit exists and is equal to zero, so \(f_{y}(0,0)\) exists. Note that the other part of the function is not defined at (0,0).

Test for differentiability

A function is differentiable at a point if the limit as \(x\) and \(y\) approaches that point is the same along any path. Here we need to check the limit along different paths to the point (0,0) and check if the rate of change is the same. Let's use the path \(y = kx\) as an example. The function becomes \(f(x, y)=f(x, kx)=\frac{5x^{2}kx}{x^{3}+k^{3}x^{3}}=\frac{5k}{1+k^{3}}\), which depends on \(k\). Therefore, the limit as \(x\) approaches to 0 is different for different values of \(k\), which proves the function is not differentiable at (0,0).

Key concepts

Partial Derivatives

When dealing with functions that have more than one variable, it’s essential to understand how the function changes with respect to each of those variables independently. This is where partial derivatives come into play. A partial derivative measures how a function changes as one variable changes, while the other variables are held constant. For instance, in the given exercise, to compute the partial derivative of the function \( f(x, y) \) with respect to \( x \), denoted as \( f_x(x,y) \), we treat \( y \) as a constant and vice versa for the partial derivative with respect to \( y \).

The existence of these partial derivatives at a point is often a first step in determining differentiability at that point. However, the existence alone does not guarantee that a function is differentiable there—this is a common misconception. The concept of a function being differentiable involves more than just these individual slices through the function landscape.

Limits and Continuity

Limits are foundational in calculus. They help us understand the behavior of functions as we approach a particular point, often where the function itself may not be defined. In multivariable calculus, taking limits gets a bit trickier as you can approach a point from an infinite number of directions in the space.

Continuity, on the other hand, is a condition that ensures a function doesn't have any abrupt changes, jumps, or holes. For a function of several variables to be continuous at a point, the limit must be the same no matter the path taken to approach that point. The exercise provided highlights that while we can find the limits of partial derivatives as we approach the origin, this alone is insufficient to prove the continuity of the function at that point. For true continuity and differentiability, the behavior must be consistent from all directions.

Multivariable Calculus

Moving beyond single-variable calculus, multivariable calculus deals with functions that have more than one input. This can get complex, as we now have to consider how a function behaves in a multidimensional space. As seen in the exercise, even if the partial derivatives exist, this multidimensional behavior may not be uniform, leading to non-differentiability.

In our example, approaching the origin along different paths yields different results, revealing the intricate dance of change across multiple dimensions. Multivariable calculus techniques, such as taking partial derivatives and evaluating limits, are tools that help us navigate this dance and understand the changing landscape of these functions. It's the interaction between these tools that paint the full picture of a function's behavior in space.

Non-Differentiable Functions

A function might not be differentiable for several reasons. Sharp corners, cusps, vertical tangents, or discontinuities all break the smoothness required for differentiability. In the context of the provided exercise, the function given by \( f(x, y) \) is shown to be non-differentiable at the origin, despite having partial derivatives defined there.

The key lies in the function's behavior as we approach the origin along different paths—in our case, the path \( y=kx \). The limit depending on the value of \( k \) indicates that the approach to the origin is inconsistent, thus violating the condition for differentiability. Non-differentiability can often be linked to a function's more complex or irregular behavior in a multidimensional space, and it's a crucial concept students should grasp to understand the limitations of calculus tools.