Question

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by \[ f(x,y) = \begin{cases} \dfrac{2x^3 + 3y^3}{x^2 + y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases} \] 

Let $f_x(0,0)$ and $f_y(0,0)$ denote first order partial derivatives of $f(x,y)$ at $(0,0)$. Which one of the following statements is TRUE? 
 

• A. $f$ is continuous at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ do not exist
• B. $f$ is differentiable at $(0,0)$
• C. $f$ is not differentiable at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ exist
• D. $f$ is not continuous at $(0,0)$ but $f_x(0,0)$ and $f_y(0,0)$ exist

Hint:

A function can have existing partial derivatives at a point yet fail to be differentiable — check using the total differential limit.

Answer 1

The Correct Option is C

Step 1: Check continuity at $(0,0)$.
Along $y=0$, \[ f(x,0) = \frac{2x^3}{x^2} = 2x \to 0. \] Along $x=0$, \[ f(0,y) = \frac{3y^3}{y^2} = 3y \to 0. \] Hence, $f$ is continuous at $(0,0)$.

Step 2: Compute partial derivatives.
\[ f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{2h - 0}{h} = 2, \] \[ f_y(0,0) = \lim_{h \to 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h \to 0} \frac{3h - 0}{h} = 3. \] Thus, partial derivatives exist.

Step 3: Check differentiability.
If $f$ were differentiable, \[ f(x,y) \approx f(0,0) + f_x(0,0)x + f_y(0,0)y = 2x + 3y. \] Compute the remainder: \[ R(x,y) = f(x,y) - (2x + 3y) = \frac{2x^3 + 3y^3}{x^2 + y^2} - 2x - 3y. \] Let $(x,y) = (t,t)$, \[ R(t,t) = \frac{5t^3}{2t^2} - 5t = -\frac{5t}{2} \neq 0 \text{ as } t \to 0. \] Hence, $\frac{R(x,y)}{\sqrt{x^2 + y^2}}$ does not approach 0, so $f$ is not differentiable.

Step 4: Conclusion.
$f$ is continuous and has partial derivatives at $(0,0)$ but is not differentiable there.