# Finding a non-differentiable function with these properties:

Wondering if anyone can point out a function $$f$$ from an open disc in $$\mathbb R^2$$ to $$\mathbb R$$, that:

• is defined and continuous at a point $$p$$;
• has(they exist) both of its partial derivatives at $$p$$;
• at least one of the partials is discontinuous at $$p$$;
• is NOT differentiable at $$p$$;

Context: (not necessary for providing an answer): While learning about the differentiablity of multivariable functions in my calculus course, started thinking about this particular case after finding out that the differentiablity of a continuous multivariable function (at a point) depends not solely on the existence of the partial derivatives. Naturally the question: “what other condition is necessary for differentiablity then?” arises. Learned that if both partials are continuous at p, that is sufficient. But apparently even if they aren't continuous at p, f can still be differentiable. Found one of those functions after much work but then wondered what would a function look like, that had everything set up for differentiablity at a point, but isn't, because of the discontinuity of one of the partials.

Would appreciate the function, or even a good intuition on how to find one.

Define $$f(r\cos(t),r\sin(t))=r^{1/2}\sin(2t)\quad(r>0,t\in\Bbb R).$$
Then $$f$$ is continuous, and not differentiable at the origin (for example if $$\phi(x)=f(x,x)=|x|^{1/2}$$ then $$\phi'(0)$$ does not exist).
But $$f(x,0)=f(0,y)=0,$$hence $$\frac{\partial f}{\partial y}(0,0)=\frac{\partial f}{\partial y}(0,0)=0.$$