If two-variable function $\ f(x,y)$ is partialy differentiable respect to both variables ; if both $\frac{\partial f}{\partial x} $ and $\frac{\partial f}{\partial y} $ exist, is it always true the second mixed partial $\frac{\partial^2 f}{\partial x \partial y} $ exist?

If it's not the general case, what could be the weakest condition that fit?

Well using my intuition $\ f_x$ and $\ f_y$ being continuous is the best I could imagine, but I guess that's not all? Please enlighten me.

I'm currently learning calculus by Stewart calculus ed.8

If exact answer requires rigorous understanding of math concepts Which books/texts do I need to look for?



1 Answer 1


Try $$f(x,y)=e^{1/(x^2+y^2)} x y, \qquad f(0,0)=0$$ Away from $(0,0)$ it is smooth, but at $(0,0)$ only the partial derivatives in directions $(1,0)$ and $(0,1)$ exist, not in the direction $(1,1)$. The $\frac{\partial^2 f}{\partial x \partial y}$ second derivative doesn't exist neither.

Another example would be

$$g(x,y)=xy\sin(\frac1{x^2+y^2})$$ It is smooth away from $(0,0)$.

$g(x,y)=O(x^2+y^2)$ so the derivatives in every directions exist and vanish at $(0,0)$, but $(\partial_x g)(0,y)=y\sin(1/y^2)$ which is not differentiable at $y=0$ so that $(\partial_y (\partial_x g))(0,0)$ doesn't exist.

  • $\begingroup$ That's appreciatable counterexample! Now I can see. $\endgroup$
    – Dan Jones
    Jan 7, 2022 at 9:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .