Solubility of an algebraic group and Weil restriction For $\ell/k$  a finite separable extension of fields and an affine variety $X_{/\ell}$, the Weil restriction of scalars $R_{\ell/k}(X_{/\ell})$ represents the functor $R_{\ell/k}(X_{/\ell})(S) := X_{/\ell}(S \times_k \ell)$, for any $k$-scheme $S$. One can show that if $X$ is a smooth affine algebraic group over $\ell$, then so is $R_{\ell/k}(X_{/\ell})$, but over $k$. Moreover, for any affine smooth algebraic group, the Weil restriction of scalars preserves finiteness, commutativity, reductivity, connectivity, unipotency, and possibly other properties. Now my question:

Does the Weil restriction of scalars preserve solubility of a smooth affine algebraic group?

I can't find any references on this. Also, I've tried showing that the Weil restriction commutes with the formation of derived subgroups, but with no success (due to my lack of ability, probably). 
For completeness, an algebraic group is soluble if its derived series becomes trivial in a finite number of steps. 
 A: [The original question did not have the separability hypothesis.  In the separable case everything is very easy, as explained in the 2nd paragraph below.  Most of this answer is about inseparable phenomena.]
Basically, you should learn about pseudo-reductive groups.  It is all about dealing with the many ways in which most of your expectations (and even things which you thought you had proved) are wrong. There is a very nice way to deal with these problems, but it takes some time to get accustomed to it. You should essentially throw out any belief that the separable case provides good guidance for the inseparable case; it does not. 
First, let's do the basic calculation to focus on where the substance lies: 
if $X$ is an affine $\ell$-scheme of finite type then ${\rm{R}}_{\ell/k}(X) \otimes_k k_s = \prod {\rm{R}}_{\ell_i/k_s}(X_i)$ where $\{\ell_i\}$ is the set of factor fields of $\ell \otimes_k k_s$ and $X_i$ is the $\ell_i$-fiber of $X \otimes_k k_s = X \otimes_{\ell} (\ell \otimes_k k_s)$.  Hence, if $\ell$ is separable (so $\ell_i = k_s$ for all $i$) then everything you ask about affirmative and quite trivial by considerations with direct products over $k_s$.  Thus, the real content, if you really did want to consider general finite extensions, is all about the case of the extensions $\ell_i/k_s$ (you may have second thoughts now), is with purely inseparable finite extensions.  
Since Weil restriction is functorially a pushforward, it is left-exact when applied to a left-exact sequence of (affine finite type) group schemes, so by preservation of commutativity we get preservation of solvability.  But basically everything else breaks down badly: finiteness is not preserved (but affineness is, so properness is ruined too, unsurprising if you think about the valuative criterion), (geometric) connectedness is not preserved away from the smooth case, the formation of derived groups also is not compatible (say in the smooth case, beyond which it isn't even clear what "derived group" should mean anyway).  Likewise, surjectivity is ruined too (for non-smooth morphisms), and reductivity is very badly destroyed.  Even being non-empty can be ruined (though not relevant in the presence of a section, as for group schemes).
Here are some instructive examples which illustrate the crazy stuff that happens.
Example 1:  Let $k$ be an imperfect field of characteristic $p > 0$ and $k'/k$ a finite nontrivial extension contained inside $k^{1/p}$.  Then the natural monomorphism between affine $k$-group schemes 
$${\rm{R}}_{k'/k}(\mu_p)/\mu_p \rightarrow {\rm{R}}_{k'/k}({\rm{GL}}_1)/{\rm{GL}}_1$$
is an isomorphism.  Indeed, since ${\rm{GL}}_1$ is $p$-divisible and the $p$-power on ${\rm{R}}_{k'/k}({\rm{GL}}_1)$ is valued in the subgroup ${\rm{GL}}_1$ (as by $k$-smoothness it suffices to check this on the Zariski-dense locus of $k_s$-points, where it becomes the fact that $(k'_s)^p \subset k_s$), by fppf-local calculation we see the map is an fppf surjective, hence an isomorphism.
This shows that ${\rm{R}}_{k'/k}(\mu_p)$ has positive dimension $[k':k]-1>0$ (but it has no non-trivial $k_s$-points since $(k'_s)^{\times}[p]=1$, so it is nowhere smooth).  In down-to-earth terms, its geometric points are the 1-units of $(k' \otimes_k \overline{k})^{\times}$, a rather huge group.  So finiteness is generally ruined.
Note also that ${\rm{R}}_{k'/k}({\rm{GL}}_1)$ modulo a one-dimensional torus is $p$-torsion (of dimension $> 0$), so this Weil restriction is not a torus (massive unipotent radical, again expressing those geometric 1-units).  In particular, this illustrates loss of reductivity (though this is a lame example since it is commutative; we will do much better in a moment).
Example 2:  Now we feed Example 1 into the exact sequence
$$1 \rightarrow \mu_p \rightarrow {\rm{SL}}_p \rightarrow {\rm{PGL}}_p \rightarrow 1$$
of $k'$-groups ($k'/k$ as above).  Applying the functor ${\rm{R}}_{k'/k} = f_{\ast}$ where $f: {\rm{Spec}}(k') \rightarrow {\rm{Spec}}(k)$, we get an exact sequence of fppf group sheaves
$$1 \rightarrow {\rm{R}}_{k'/k}(\mu_p) \rightarrow {\rm{R}}_{k'/k}({\rm{SL}}_p) \rightarrow {\rm{R}}_{k'/k}({\rm{PGL}}_p) \rightarrow {\rm{R}}^1(f_{\ast})(\mu_p).$$
So by dimension consideration and smoothness stuff we see two things:  ${\rm{R}}_{k'/k}({\rm{PGL}}_p)$ contains ${\rm{R}}_{k'/k}({\rm{SL}}_p)/{\rm{R}}_{k'/k}(\mu_p)$ as a smooth normal closed subgroup of codimension $[k':k] - 1 > 0$ for which the cokernel (necessarily nontrivial) is commutative!  Hence, ${\rm{R}}_{k'/k}({\rm{PGL}}_p)$ is not its own derived group (in fact its derived group is the image of ${\rm{R}}_{k'/k}({\rm{SL}}_p)$, the latter being perfect ultimately because ${\rm{SL}}_p$ is simply connected), and we have exhibited an (inseparable) isogeny whose Weil restriction is not surjective!  
Observe that ${\rm{R}}_{k'/k}({\rm{SL}}_p)/\mu_p$ contains the central unipotent smooth connected subgroup ${\rm{R}}_{k'/k}(\mu_p)/\mu_p \ne 1$ from Example 1, so ${\rm{R}}_{k'/k}({\rm{SL}}_p)$ is not reductive (that central unipotent smooth connected subgroup is not the entire "geometric" unipotent radical; to explain the discrepancy between the $k$-structure and the unipotent radical over $\overline{k}$ is one of the main difficulties in setting up the theory of pseudo-reductive groups).
Example 3: Under a smoothness hypothesis, geometric connectedness is preserved.  However, this is not at all obvious (think about it!).  See Proposition A.5.9 of the book "Pseudo-reductive groups" for a proof (without group structure, which is irrelevant).  In fact, the smoothness hypothesis is absolutely essential: over any imperfect field $k$ one can make geometrically integral affine curves which are smooth away from one point but whose Weil restriction is disconnected with a connected component that is nowhere reduced!  Over some $k$ one can even arrange the non-smooth point to be a regular point.  Crazy stuff.
Example 4: Let $k'/k$ be as in Example 1 and let $A'$ be a nonzero $k'$-algebra of finite type.  When is ${\rm{R}}_{k'/k}(A')$ non-empty?  The condition is that it has to have a $\overline{k}$-point, which is to say that $k' \otimes_k A'$ admits a $k' \otimes_k \overline{k}$-point over $k'$.  Let $k' = k(a^{1/p})$ for $a \in k - k^p$ and $A' = k(a^{1/p^2})$, so we need to find a $k$-algebra map $k(a^{1/p^2}) \rightarrow \overline{k} \otimes_k k' = \overline{k}[t]/(t^p)$ under which $a^{1/p} \in k''$ is sent to $a^{1/p} + t$.  But $a^{1/p}$ has a $p$th root in $k(a^{1/p^2})$ yet the element $a^{1/p} + t \in \overline{k}[t]/(t^p)$ does not have a $p$th root (as all $p$th powers in this algebra lie in $\overline{k}$).  Thus, in this case the Weil restriction is empty.
There are further kinds of disorienting behavior that arise in connection with inseparable Weil restriction, but the above seems sufficient to give you due caution when thinking about these matters.  Poke around in the book "Pseudo-reductive groups" for more examples and discussion.
A: Let me try to deduce a positive answer from standard properties of the Weil restriction and the results presented in this paper (found by Google): http://math.stanford.edu/~conrad/papers/appbnew.pdf. This is probably an overkill, especially if you are only interested in perfect ground fields.
Weil restriction preserves left exactness; by Corollary A.5.4 (1) of "Pseudo-reductive groups" by Conrad, Gabber, Prasad, it also preserves smooth surjections between affine groups. Thus, all you need to prove is that your smooth affine $\ell$-group $X$ has a composition series whose subquotients are all affine smooth and commutative. In the sequel, we argue the existence of such a filtration. For this, as a first reduction, we may assume that $X$ is connected (its component group is etale and inherits solvability).
Since tori are affine smooth and commutative, Theorem 5.4 in the linked pdf reduces us to assuming that $X$ is unipotent and either $k$-split or $k$-wound. The split case is immediate from Definition 1.3. In the $k$-wound case, the filtration by cckp-kernels provided by Corollary 3.3 gives the claim.
EDIT: The original question did not have the assumption that $\ell/k$ be separable.
