Question

Let 𝑓:ℝ×ℝ→ℝ be a function defined by
\(f(x,y)=\begin{cases} \frac{xy(x+y}{x^2+y^2} & \quad \text{if }(x,y)≠(0,0),\\  0, & \quad if\,(x,y)=(0,0). \end{cases}\)
Then, which of the following statements is/are TRUE?

• A. 𝑓 is continuous on ℝ × ℝ
• B. The partial derivative of 𝑓 with respect to 𝑦 exists at (0, 0), and is 0
• C. The partial derivative of 𝑓 with respect to 𝑥 is continuous on ℝ × ℝ
• D. 𝑓 is NOT differentiable at (0, 0)

Answer 1

The Correct Option is A, B, D

To solve this problem, we need to examine different properties of the function \( f(x, y) = \frac{xy(x+y)}{x^2 + y^2} \) when \((x, y) \neq (0, 0)\) and \(f(0, 0) = 0\). Let's explore each option one by one:

1. Consider the continuity of \(f\) at \((0, 0)\). For \(f\) to be continuous at \((0, 0)\), the limit of \(f(x, y)\) as \((x, y) \to (0, 0)\) must be \(0\), which is the value of \(f\) at the origin.

We compute:

\[ \lim_{(x,y) \to (0,0)} \frac{xy(x+y)}{x^2+y^2} \]

Substituting polar coordinates \(x = r \cos \theta\) and \(y = r \sin \theta\), we get:

\[ \frac{r^2 \cos \theta \sin \theta (r \cos \theta + r \sin \theta)}{r^2 \cos^2 \theta + r^2 \sin^2 \theta} = \frac{r^3 \cos \theta \sin \theta (\cos \theta + \sin \theta)}{r^2} = r \cos \theta \sin \theta (\cos \theta + \sin \theta) \]

As \(r \to 0\), this expression tends to 0 regardless of \(\theta\), so \(f\) is continuous everywhere, including at \((0,0)\).

2. The partial derivative of \(f\) with respect to \(y\) at \((0, 0)\) is:

\[ \frac{\partial f}{\partial y} (0, 0) = \lim_{h \to 0} \frac{f(0, h) - f(0, 0)}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \]

Thus, the partial derivative with respect to \(y\) exists at \((0,0)\) and is 0.

3. To check if \(f\) is differentiable at \((0, 0)\), it must have partial derivatives at \((0, 0)\) and satisfy the condition

\[ \lim_{(x, y) \to (0,0)} \frac{f(x, y) - f(0, 0) - A x - B y}{\sqrt{x^2 + y^2}} = 0 \]

After finding that \(A = 0\) and \(B = 0\), substitute back, we have:

\[ \lim_{(x, y) \to (0,0)} \frac{xy(x+y)}{(x^2 + y^2)^{3/2}} \]

This limit depends on the path taken, indicating \(f\) is not differentiable at \((0, 0)\).

Thus, the correct answers are:

𝑓 is continuous on ℝ × ℝ

The partial derivative of 𝑓 with respect to 𝑦 exists at (0, 0), and is 0

𝑓 is NOT differentiable at (0, 0)