Question
Show that the function f defined by
is not differentiable at (0,0). Does the partial derivatives of f exists at (0,0)? or any at any other point in R²? Justify your answer.
Answer :
Word Count : 543
We are given the function: \[ f(x, y) = \begin{cases} \frac{xy}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0). \end{cases} \] We are asked to prove that this function is not differentiable at (0, 0) and whether the partial derivatives exist at \((0, 0)\) or at any other point in \(\mathbb{R}^2\). ### Step 1: Check if the partial derivatives exist at \((0, 0)\) The partial derivatives at \((0, 0)\) are given by the limits: \[ \frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}, \quad \frac{\partial f}{\partial y}(0, 0) = \lim_{h \to ________ __________ ____ ____ ___ _____ ______.
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We are given the function: \[ f(x, y) = \begin{cases} \frac{xy}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0). \end{cases} \] We are asked to prove that this function is not differentiable at (0, 0) and whether the partial derivatives exist at \((0, 0)\) or at any other point in \(\mathbb{R}^2\). ### Step 1: Check if the partial derivatives exist at \((0, 0)\) The partial derivatives at \((0, 0)\) are given by the limits: \[ \frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}, \quad \frac{\partial f}{\partial y}(0, 0) = \lim_{h \to ________ __________ ____ ____ ___ _____ ______.
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