Part CCSIR NET June 2024partials-exist-without-continuity

Partials exist without continuity

Define $f : R^{2} \to R$ by $f (x, y) = y x^{2} + y^{2} / x$ if x ≠ 0, and f(x, y) = 0 if x = 0. Which of the following statements are true?

A. $\partial f / \partial x (0, 0)$ exists✓
B. $\partial f / \partial y (0, 0)$ exists✓
C.f is not continuous at (0, 0)✓
D.f is not differentiable at (0, 0)✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

50 are analysed free — try those first.

Related counterexample: If all partial derivatives exist at a point then f is continuous there

More on this topic

Part C questionDecember 2023
monotone not necessarily increasingDecember 2023
projection is open mapDecember 2023
differentiability costs one more power than continuityDecember 2024
oddness makes the restriction linear not just positively homogeneousDecember 2024
partials exist without differentiabilityDecember 2024

The chapter behind this: Differentiability in several variables — free to read

From Functions of Several Variables › Partial derivatives, differentiability, chain rule

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