Let $(\Omega) \subset \mathbb{R}.$ Suppose $f\in W^{k,p}(\Omega)$ then its first order weak derivatives (which is a distributional derivative) belongs to $W^{k-1,p}(\Omega)$ for $k\geq 1.$
Suppose $f\in L^p(\Omega).$ Does the first order distributional derivative of $f$ belong to $W^{-1,p}(\Omega)?$ If so how to prove it?
Similarly, if $f\in W^{-k,p}(\Omega).$ Does the first order partial derivative belong to $f\in W^{-k-1,p}(\Omega)?$
P.S: $W^{-k,p}(\Omega)$ is defined as the dual of $W_0^{k,q}(\Omega).$
A problem of definition? If $\phi \mapsto \langle f,\phi \rangle\in W^{k,q}_0(\Omega)'$ then $\phi \mapsto \langle f,\phi' \rangle\in W^{k+1,q}_0(\Omega)'$
Its restriction to $\phi \in C^\infty_c(\Omega)$ gives the distribution $\phi \mapsto \langle f,\phi' \rangle\in D'(\Omega)$ that we are allowed to denote $\phi \mapsto -\langle f',\phi \rangle$
and which is the distributional derivative of $\phi \mapsto -\langle f,\phi \rangle\in D'(\Omega)$.
So yes the distributional derivative of the distribution represented by $f\in W^{k,q}_0(\Omega)'$ extends to an element of $W^{k+1,q}_0(\Omega)'$.
Is there another interpretation of your question?