Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be defined by \[ f(x, y) = \begin{cases} \frac{(x^2 - y^2)^2 xy}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0). \end{cases} \] Then, the value of \( \frac{\partial f}{\partial y}(0, 0) \) and \( \frac{\partial f}{\partial x}(0, 0) \) is equal to ........... (rounded off to two decimal places).

Hint:

When computing partial derivatives at a point, use the definition of the derivative as a limit and be careful with cases where the function is piecewise defined.

Answer 1

Step 1: Compute \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial x} \).
For \( f(x, y) \), we need to compute the partial derivatives at the point \( (0, 0) \). We apply the definition of partial derivatives: \[ \frac{\partial f}{\partial y}(0, 0) = \lim_{h \to 0} \frac{f(0, h) - f(0, 0)}{h}, \quad \frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}. \] Evaluating these limits using the given expression for \( f(x, y) \), we find: \[ \frac{\partial f}{\partial y}(0, 0) = 0, \quad \frac{\partial f}{\partial x}(0, 0) = 0. \] Final Answer: \[ \boxed{0}. \]