the space $\frac d{dx} W^{2, s}$ contained in $W^{2,s-1}$? If we take the domain is Sobolev space $W^{2,s}=\{f: \frac {d^\alpha} {dx^\alpha}f \in L^2, \forall 0\leq\alpha \leq s\}.$ Consider the operator $\frac d{dx}$. What is the image space $\frac d{dx}W^{2, s}=\{\tfrac{df}{dx}(x): f\in W^{2, s}\}$?  Is it contained in $W^{2,s-1}$?
If we take the domain $L^2$, What is the image space $\frac d{dx}L^2=\{\tfrac{df}{dx}(x): f\in L^2\}$? Is it contained in $W^{2,1}$?
 A: First of all I think it is not very clear what you mean by $\tfrac{d}{dx}W^{2,s}$. If I am not wrong, by this you mean any function of the form $\tfrac{df}{dx}(x)$, where $f\in W^{2,s}$. If that is the case, then no, the space $\tfrac{d}{dx}L^2$ is not contained in $W^{2,1}$. For example, consider the function $$
f(x)=e^{-x^2}H(x),
$$
where $H(x)$ is the Heaviside function centered at $x=0$. It is not difficult to see that $\tfrac{df}{dx}$ contains a Dirac delta at the origin, and hence $\tfrac{df}{dx}$ doesn't belong to $W^{2,1}$, not even to $L^2$. Of course, notice that $f\in L^2$.
To answer your first question, yes, any function $\tfrac{df}{dx}$ with $f\in W^{2,s}$ belongs to $W^{2,s-1}$. This follows directly from taking the $W^{2,s-1}$-norm. However, here is the problematic point of what you mean by $\tfrac{d}{dx}W^{s,2}$. For example, constant functions $f(x)\equiv c$ satisfy that $$
\tfrac{df}{dx}\equiv 0 \in W^{2,s},
$$
although they cannot be represented as $\tfrac{dg}{dx}(x)$ for some $g\in W^{2,s}$ (at least not if the domain is the whole line $\mathbb{R}$). I hope this helps you.
