Suppose $u$ is a tempered distibution in $\mathbb{R}^n$. How are supp$(\widehat{u})$ and support of $\sum_{|\alpha|\leq k}\frac{\partial^{\alpha}\widehat{u^n}}{\partial x^{\alpha}}$ compared , where $n$ is a positive integer.
$\textbf{edit:}$ Do we have the inclusion supp$(\sum_{|\alpha|\leq k}\frac{\partial^{\alpha}\widehat{u^n}}{\partial x^{\alpha}})\subseteq$ supp$(\widehat{u})$?
First of all, note that $u^n$ is not well-defined for tempered distributions, because you can't multiply them. Therefore we need suppose that $u$ is function.
One of extreme cases is $u= x^p$ for $p\in \Bbb N$. Easy to see that $\sum_{|\alpha|\leq k}\frac{\partial^{\alpha}\widehat{u^n}}{\partial x^{\alpha}}$ will be supported in the point $\{0\}$.
Another case is $u(x) = \exp\left(-\frac{x^2}{2}\right)$ (a distribution $N_{0,1}$), then its Fourier transform is of the form $N_{0,1}$ (up to some irrelevant constants related to the definition of Fourier transform).
Therefore, $$\sum_{|\alpha|\leq k}\frac{\partial^{\alpha}\widehat{u^n}}{\partial x^{\alpha}}$$ for $n=1$ is supported in $\Bbb R$.
Finally, you can take $u \in \mathcal S$ such that $\widehat {u^n} = N_{0,1}$ (such $u$ exists because Fourier transform is the continuously invertible operator on the space of Schwartz functions). Easy to see that the support of the sum will be still $\Bbb R$.
edit another example for $n=1$: let $$\hat u=\exp(-|x|),\quad \text{supp } \hat u = \Bbb R.$$ It is a tempered distribution. Then $$\hat u' =\begin{cases}e^x,&x\le 0,\\-e^{-x},&x>0,\end{cases} $$ therefore the support of $\hat u+\hat u'$ is $(-\infty,0]$, i.e. it decreased.
What you can't achieve is to strictly increase the support for $n=1$. I think one can easily tinker the above examples to show that support of the sum is the subset of the support of $\hat u$ for all $n\ge 1$.