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Part CCSIR NET December 2024partials-exist-without-differentiability

Partials exist without differentiability

For a positive real number a,aa, \sqrt{a} denotes the positive square root of a. Consider the function f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R} defined by f(x, y) = (x/|x|)x2+y2)\sqrt{x^{2} + y^{2}} for x ≠ 0, and f(x, y) = 0 for x = 0. Which of the following statements are true?

  1. A.f is continuous at (0, 0).✓ Correct
  2. B.The partial derivatives ∂f/∂x\partial{}f/\partial{}x and ∂f/∂y\partial{}f/\partial{}y exist at (0, 0).✓ Correct
  3. C.f is differentiable at (0, 0).
  4. D.f is not differentiable at (0, 0).✓ Correct

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, and why this one tests standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

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

More on this topic

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

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

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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