Question

Let \(f: \mathbb{R}^2 \to \mathbb{R}\) be a function defined as follows:
\[ f(x,y) = \begin{cases} \dfrac{x\sqrt{x^2+y^2}}{|x|} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases} \]
Which of the following statements is/are TRUE?

• A. \(f(x,y)\) is continuous at \((0,0)\).
• B. For any \(u \in \mathbb{R}^2\), the directional derivative of \(f\) at \((0,0)\) in the direction of \(u\) exists.
• C. \(\dfrac{\partial f}{\partial x}(0,0) = 1 = \dfrac{\partial f}{\partial y}(0,0)\).
• D. \(f(x,y)\) is differentiable at \((0,0)\).

Hint:

Rewrite \(f(x,y) = \text{sign}(x)\sqrt{x^2+y^2}\) for \(x \neq 0\) and check whether this bound goes to zero along every path, and whether the directional derivative is linear in direction.

Answer 1

The Correct Option is A, B

Step 1: Rewrite the function in a simpler form.
For \(x \neq 0\), \(\dfrac{x}{|x|}\) equals \(1\) when \(x>0\) and \(-1\) when \(x<0\); call this \(\text{sign}(x)\). So for \(x \neq 0\),
\[ f(x,y) = \text{sign}(x)\sqrt{x^2+y^2} \]
and \(f(0,y) = 0\) for every \(y\). This form makes it much easier to check each property at the origin.

Step 2: Check continuity at \((0,0)\), option (A).
For every \((x,y)\), whether \(x\) is zero or not, we have
\[ |f(x,y)| \leq \sqrt{x^2+y^2} \]
because \(|\text{sign}(x)| = 1\) when \(x \neq 0\), and \(f(0,y)=0\) matches the bound too. As \((x,y) \to (0,0)\) along any path, the right side \(\sqrt{x^2+y^2} \to 0\), so by the squeeze theorem \(f(x,y) \to 0 = f(0,0)\). This holds along every path, so \(f\) is continuous at \((0,0)\). Option (A) is TRUE.

Step 3: Check the directional derivative for a general direction, option (B).
Take any unit vector \(u = (u_1,u_2)\) and look at \(g(t) = f(tu_1,tu_2)\) for small \(t>0\).
If \(u_1 \neq 0\): \(g(t) = \text{sign}(tu_1)\sqrt{t^2u_1^2+t^2u_2^2} = \text{sign}(u_1)\cdot t\sqrt{u_1^2+u_2^2} = \text{sign}(u_1)\cdot t\) (since \(u\) is a unit vector). So
\[ \lim_{t\to0^+}\frac{g(t)-g(0)}{t} = \text{sign}(u_1) \]
which is a finite number. If \(u_1 = 0\) (direction along the \(y\)-axis), then \(g(t) = f(0,tu_2) = 0\) for every \(t\), so the derivative is \(0\). Either way, the directional derivative exists and is finite for every direction \(u\). Option (B) is TRUE.

Step 4: Check the partial derivatives, option (C).
Using Step 3 with \(u=(1,0)\): \(\dfrac{\partial f}{\partial x}(0,0) = \text{sign}(1) = 1\).
Using Step 3 with \(u=(0,1)\): since \(u_1=0\), \(\dfrac{\partial f}{\partial y}(0,0) = 0\).
So \(\dfrac{\partial f}{\partial x}(0,0)=1\) but \(\dfrac{\partial f}{\partial y}(0,0)=0\), not \(1\). Option (C) claims both equal \(1\), so (C) is FALSE.

Step 5: Check differentiability, option (D).
If \(f\) were differentiable at \((0,0)\), the directional derivative in direction \(u=(u_1,u_2)\) would have to equal the linear expression \(\nabla f(0,0)\cdot u = 1\cdot u_1 + 0\cdot u_2 = u_1\). But Step 3 showed the actual directional derivative is \(\text{sign}(u_1)\), which is \(+1\) or \(-1\) and does not vary smoothly with \(u_1\) the way a linear function does. For instance, take \(u = (\tfrac{1}{\sqrt2},\tfrac{1}{\sqrt2})\): the linear guess gives \(\tfrac{1}{\sqrt2} \approx 0.707\), but the true directional derivative is \(\text{sign}(u_1)=1\). Since these do not match, \(f\) is not differentiable at \((0,0)\). Option (D) is FALSE.

Final Answer:
\(f\) is continuous at the origin and every directional derivative there exists, so (A) and (B) hold. But the two partial derivatives are not both \(1\), and \(f\) is not differentiable, so (C) and (D) fail.
\[ \boxed{\text{(A) and (B)}} \]