Counterexample

Is this true?

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

No — it is false.

The counterexample

f(x,y) = xy $/(x^{2}+y^{2}), f(0,0) = 0$

Why it fails

Both partials are 0 at the origin, but f = ½ along y = x, so f is not continuous.

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Execution slip

The idea was right. The computation was not.

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Go deeper· Explained in full in Differentiability in several variables· Asked in CSIR NET June 2025· From Functions of Several Variables › Partial derivatives, differentiability, chain rule

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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