Question

Let \( f(x, y) = |xy| + x \) for all \( (x, y) \in \mathbb{R}^2 \). Determine the existence of the partial derivative of \( f \) with respect to \( x \) exists:

• A. at \( (0,0) \) but not at \( (0,1) \).
• B. at \( (0,1) \) but not at \( (0,0) \).
• C. at both \( (0,0) \) and \( (0,1) \).
• D. neither at \( (0,0) \) nor at \( (0,1) \).

Answer 1

The Correct Option is A

1. At \( (0, 0) \): - For partial derivative \( \frac{\partial f}{\partial x} \), compute: \[ f_x(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}. \] - Substituting \( f(h, 0) = |h \cdot 0| + h = h \) and \( f(0, 0) = 0 \): \[ f_x(0, 0) = \lim_{h \to 0} \frac{h}{h} = 1. \] - The partial derivative exists at \( (0, 0) \). 2. At \( (0, 1) \): - For partial derivative \( \frac{\partial f}{\partial x} \), compute: \[ f_x(0, 1) = \lim_{h \to 0} \frac{f(h, 1) - f(0, 1)}{h}. \] - Substituting \( f(h, 1) = |h \cdot 1| + h = |h| + h \) and \( f(0, 1) = 0 \): \[ f_x(0, 1) = \lim_{h \to 0} \frac{|h| + h}{h}. \] - As \( h \to 0^+ \), \( f_x(0, 1) \to 2 \); as \( h \to 0^- \), \( f_x(0, 1) \to 0 \). - The limit does not exist, so the partial derivative does not exist at \( (0, 1) \).