Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be a function defined by \[ f(x, y) = \begin{cases} \frac{x^2 y}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0). \end{cases} \] Find the value of \( \frac{\partial f}{\partial x} \) at \( (0, 0) \).

• A. 0
• B. 1
• C. -1
• D. Undefined

Hint:

To find the partial derivative at a point where the function is piecewise defined, check the limit carefully and confirm continuity.

Answer 1

The Correct Option is B

To find \( \frac{\partial f}{\partial x} \) at \( (0, 0) \), we first check if the function is continuous at \( (0, 0) \). The function is defined as: \[ f(x, y) = \frac{x^2 y}{x^2 + y^2} \text{ for } (x, y) \neq (0, 0). \] At \( (0, 0) \), we need to check if the limit of \( f(x, y) \) as \( (x, y) \to (0, 0) \) exists: \[ \lim_{(x, y) \to (0, 0)} \frac{x^2 y}{x^2 + y^2} = 0. \] Thus, the function is continuous at \( (0, 0) \). Next, we compute the partial derivative \( \frac{\partial f}{\partial x} \) at \( (0, 0) \): \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}. \] Since \( f(h, 0) = 0 \) for all \( h \), we have: \[ \frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{0 - 0}{h} = 1. \] Thus, \( \frac{\partial f}{\partial x} = 1 \) at \( (0, 0) \).