Let $N$ be a normal subgroup of $G$ s.t. $N$ and $G/N$ are solvable, then $G$ is solvable.
Proof:
Because $N$, $G/N$ are solvable $\Rightarrow$ $N^{(s)}=\{e\}$, $(G/N)^{(t)}=\{e\}$ for $s,t\in \mathbb{N}$. Since $G/N$ is a homomorphic image of $G$ (under $\psi: G\rightarrow G/N$), we have that $(G/N)^{(n)}=G^{(n)}N/N$ for every $n\in \mathbb{N}$.
Thus, $G^{(t)}\subseteq N$ and therefore $G^{(s+t)}\subseteq N^{s}=\{e\}$ and $G$ is solvable.
I don't understand the part where $(G/N)^{(n)}=G^{(n)}N/N$ and I also don't get why we can say $G^{(t)}\subseteq N$.
I hope someone can help clearing this up for me.
For original source, follow this link (Theorem 2.1)
Maybe you should proceed by a small example: let's start with the first derivative, namely $(G/N)'=G'N/N$. How do you prove this? Well just write it out: $$(G/N)'=\langle [\overline{x},\overline{y}]: \overline{x},\overline{y} \in G/N \rangle \\=\langle [x,y]N: x,y \in G \rangle \\=\langle [x,y]: x,y \in G \rangle N \\=G'N/N.$$ And $G'N/N=\overline{1}$ if and only if $G'N=N$ if and only if $G' \subseteq N$. Now this should help you for the higher derivatives.