Question

Let \( f: \mathbb{R}^2 \to \mathbb{R} \) be a function defined by \[ f(x,y) = \begin{cases} \dfrac{y^3}{x^2 + y^2}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0) \end{cases} \] Let \( f_x(x,y) \) and \( f_y(x,y) \) denote the first-order partial derivatives of \( f(x,y) \) with respect to \( x \) and \( y \) respectively. Then, which of the following statements is FALSE?

• A. \( f_x(x,y) \) exists and is bounded at every \( (x,y) \in \mathbb{R}^2 \)
• B. \( f_y(x,y) \) exists and is bounded at every \( (x,y) \in \mathbb{R}^2 \)
• C. \( f_y(0,0) \) exists and \( f_y(x,y) \) is continuous at (0,0)
• D. \( f \) is NOT differentiable at (0,0)

Hint:

To check differentiability, ensure both partial derivatives exist and are continuous at the point. Discontinuity implies non-differentiability.

Answer 1

The Correct Option is C

Step 1: Compute partial derivatives for \( (x,y) \ne (0,0) \).
\[ f_x(x,y) = \frac{\partial}{\partial x}\left(\frac{y^3}{x^2 + y^2}\right) = \frac{-2x y^3}{(x^2 + y^2)^2} \] \[ f_y(x,y) = \frac{\partial}{\partial y}\left(\frac{y^3}{x^2 + y^2}\right) = \frac{3y^2(x^2 + y^2) - 2y^4}{(x^2 + y^2)^2} = \frac{y^2(3x^2 + y^2)}{(x^2 + y^2)^2} \]
Step 2: Evaluate at (0,0).
\[ f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = 0 \] \[ f_y(0,0) = \lim_{h \to 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h \to 0} \frac{h^3 / h^2}{h} = 1 \]
Step 3: Check continuity of \( f_y(x,y) \) at (0,0).
Along the line \( x = 0 \): \( f_y = 1 \). Along the line \( y = 0 \): \( f_y = 0 \). Hence, \( f_y(x,y) \) is not continuous at (0,0).
Step 4: Differentiability.
Since partial derivatives exist but are not continuous at (0,0), \( f \) is not differentiable at (0,0). Final Answer: \[ \boxed{(C)} \]