Question

Let the function \( f(x, y) \) be defined as \[ f(x, y) = \begin{cases} \frac{y}{|y|} \sqrt{2x^2 + 3y^2}, & y \neq 0 \\ 0, & y = 0 \end{cases} \] Then \( \frac{\partial f}{\partial y} (0, 0) \) is equal to

• A. \( \sqrt{2} \)
• B. \( \sqrt{3} \)
• C. 0
• D. 1

Hint:

For piecewise functions, ensure continuity and differentiability before computing partial derivatives. In cases involving absolute values or sign functions, consider limiting behavior.

Answer 1

The Correct Option is B

We are given the function \( f(x, y) \). We need to compute the partial derivative \( \frac{\partial f}{\partial y} (0, 0) \).

Step 1: Check for continuity and differentiability at \( (0,0) \).
To compute the partial derivative at \( (0, 0) \), we need to evaluate the limit: \[ \frac{\partial f}{\partial y} (0, 0) = \lim_{h \to 0} \frac{f(0, h) - f(0, 0)}{h}. \] For \( y \neq 0 \), we have: \[ f(0, h) = \frac{h}{|h|} \sqrt{3h^2} = \sqrt{3} \text{ sign}(h). \] Thus, \[ \frac{\partial f}{\partial y} (0, 0) = \lim_{h \to 0} \frac{\sqrt{3} \text{ sign}(h)}{h} = \sqrt{3}. \]

Final Answer: \[ \boxed{\text{(B) } \sqrt{3}}. \]