Suppose we have a non tempered distribution $u\in \mathcal D'(\mathbb R^d)\backslash \mathcal S'(\mathbb R^d)$. Is it possible to have $\partial_{x_1}...\partial_{x_d}u \in \mathcal S'(\mathbb R^d)$ where the derivative is taken in the sense of Schwartz distributions? I cannot find an example nor prove the converse. Any reference to suggest?
$e^x+e^y$ defines an obviously non tempered distribution on $\mathbb{R}^2$, whose mixed derivative $\partial_x\partial_y(e^x+e^y)=0$ ... 


A distribution is tempered if and only if it is the distributional derivative $D^nF$ of a continuous function $F$ which is $O(x^k)$ for some positive integer. Its primitive is then $D^{n1}F$ and so also tempered. Hence the answer to your question is no. (Suitable reference: J. Sebastiao e Silva, Integrals and orders of growth of distributions, Lisbon, 1964). Edit. I wrote up the onedimensional case dor simplicity. If $n=0$, you have a continous function and you simply take its classical primitive which is clearly a tempered distribution. 

