Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be given by

\[ f(x, y) = \begin{cases} \dfrac{x^2 y (x - y)}{x^2 + y^2}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases} \]

Then

\[ \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y}  \right) - \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x}  \right) \text{ at the point } (0, 0) \text{ is ..........} \]

Hint:

When computing mixed partial derivatives, always check for continuity and differentiability at the point of interest, especially when dealing with piecewise functions.

Answer 1

Correct Answer: 1

1. Compute $\frac{\partial f}{\partial x}$ at $(0,0)$

$$\frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}$$

Using the definition of $f(x,y)$:

$$f(h, 0) = \frac{h^2 \cdot 0 \cdot (h-0)}{h^2 + 0^2} = 0$$

$$f(0, 0) = 0$$

$$\frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0} \frac{0 - 0}{h} = \mathbf{0}$$

2. Compute $\frac{\partial f}{\partial y}$ at $(0,0)$

$$\frac{\partial f}{\partial y}(0,0) = \lim_{k \to 0} \frac{f(0, k) - f(0, 0)}{k}$$

Using the definition of $f(x,y)$:

$$f(0, k) = \frac{0^2 \cdot k \cdot (0-k)}{0^2 + k^2} = 0$$

$$f(0, 0) = 0$$

$$\frac{\partial f}{\partial y}(0,0) = \lim_{k \to 0} \frac{0 - 0}{k} = \mathbf{0}$$

3. Compute $\frac{\partial^2 f}{\partial y \partial x}(0,0)$

This is the derivative of $\frac{\partial f}{\partial x}$ with respect to $y$, evaluated at $(0,0)$.

$$\frac{\partial^2 f}{\partial y \partial x}(0,0) = \lim_{k \to 0} \frac{\frac{\partial f}{\partial x}(0, k) - \frac{\partial f}{\partial x}(0, 0)}{k}$$

We already found $\frac{\partial f}{\partial x}(0, 0) = 0$. Now we need $\frac{\partial f}{\partial x}(0, k)$ for $k \neq 0$.

For $(x,y) \neq (0,0)$:

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

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

Evaluate $\frac{\partial f}{\partial x}$ at $(0, k)$ where $k \neq 0$:

$$\frac{\partial f}{\partial x}(0, k) = \frac{(0 - 0)(0 + k^2) - (0 - 0)(0)}{(0 + k^2)^2} = \frac{0}{k^4} = 0$$

Substitute back into the limit:

$$\frac{\partial^2 f}{\partial y \partial x}(0,0) = \lim_{k \to 0} \frac{0 - 0}{k} = \mathbf{0}$$

4. Compute $\frac{\partial^2 f}{\partial x \partial y}(0,0)$

This is the derivative of $\frac{\partial f}{\partial y}$ with respect to $x$, evaluated at $(0,0)$.

$$\frac{\partial^2 f}{\partial x \partial y}(0,0) = \lim_{h \to 0} \frac{\frac{\partial f}{\partial y}(h, 0) - \frac{\partial f}{\partial y}(0, 0)}{h}$$

We already found $\frac{\partial f}{\partial y}(0, 0) = 0$. Now we need $\frac{\partial f}{\partial y}(h, 0)$ for $h \neq 0$.

For $(x,y) \neq (0,0)$:

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

Evaluate $\frac{\partial f}{\partial y}$ at $(h, 0)$ where $h \neq 0$:

$$\frac{\partial f}{\partial y}(h, 0) = \frac{(h^3 - 0)(h^2 + 0) - (0 - 0)(0)}{(h^2 + 0)^2} = \frac{h^5}{h^4} = h$$

Substitute back into the limit:

$$\frac{\partial^2 f}{\partial x \partial y}(0,0) = \lim_{h \to 0} \frac{h - 0}{h} = \lim_{h \to 0} 1 = \mathbf{1}$$

5. Final Calculation

$$\frac{\partial^2 f}{\partial x \partial y}(0,0) - \frac{\partial^2 f}{\partial y \partial x}(0,0) = 1 - 0 = \mathbf{1}$$