The function might be the world’s most popular example demonstrating that the existence of partial derivatives does not imply differentiability.

But in my opinion, it is somewhat extreme and potentially confusing, with discontinuity added to the mix. I prefer

pictured below.

This one is continuous. In fact, it is Lipschitz continuous because the first-order partials and are bounded. The restriction of to the line is , which is a familiar single-variable example of a nondifferentiable function.

To unify the analysis of such examples, let . Then

With , where , we get

By symmetry, is bounded as well.

My favorite example from this family is more subtle, with a deceptively smooth graph:

The formula is

Since decays almost quadratically near the origin, it is differentiable at . Indeed, the first order derivatives and are continuous, as one may observe using above.

And the second-order partials and are also continuous, if just barely. Indeed,

Since the growth of is sub-logarithmic, it follows that and . Hence,

So, as . Even though the graph of looks quite similar to the first example in this post, this one is continuous. Can’t trust these plots.

By symmetry, is continuous as well.

But the **mixed partial does not exist** at , and tends to as . The first claim is obvious once we notice that and blows up at . The second one follows from

where while the other two terms tend to zero, as in the estimate for . Here is the graph of .

This example is significant for the theory of partial differential equations, because it shows that a solution of the Poisson equation with continuous may fail to be in (twice differentiable, with continuous derivatives). The expected gain of two derivatives does not materialize here.

The situation is rectified by upgrading the continuity condition to Hölder continuity. Then indeed gains two derivatives: if for some , then . In particular, the Hölder continuity of and implies the Hölder continuity of .