Prove that if both $N$ and $G/N'$ are nilpotent, then $G$ is nilpotent. Let $G$ be a group, and let $N$ be a normal subgroup of $G$. Let $N'$ be the derived group of $N$. Prove that if both $N$ and $G/N'$ are nilpotent, then $G$ is nilpotent. Furthermore, assume that $G$ is arbitrary, not necessarily finite. Prove that if $N$ is nilpotent and $G/N'$ is abelian, thn $G$ is nilpotent and $C_{i}(G)=C_{i}(N)$ for $2 \leq i $
Recall that $C_{i}(G)$ is the higher commutator subgroup of a group $G$
I think the lemma maybe usefull: If the factor group $G/ \Phi G$ is nilpotent for a finite group $G$, then $G$ itself is nilpotent.
So I think we should prove that if $G/N'$ is nilpotent then $G/ \Phi G$ is nilpotent.
How do I prove it?
P/s: There is a details proof on the book The Theory of Infinite Soluble Groups, JOHN C. LENNOX and DEREK J. S. ROBINSON, $(1.2.17)$ page $12$, but it looks complicated because they use stronger tools
 A: For finite groups there is an easier proof, using the following facts. The proof of these facts can be found in the spoiler. All groups here are finite. Recall that $\Phi(G)$ is the set of non-generators of $G$ and the intersection of all maximal subgroups of $G$. And that a group is nilpotent if and only if all its maximal subgroups are normal (here is where we need finiteness).
Lemma 1 A group $G$ is nilpotent if and only if $G' \subseteq \Phi(G)$.
Lemma 2 If $N \unlhd G$, then $\Phi(N) \unlhd \Phi(G)$.
Now assume that $G/N'$ and $N$ are nilpotent. By Lemma 1, we have $N' \subseteq \Phi(N)$, and by Lemma 2, it follows that $N' \unlhd \Phi(G)$. If $M$ is a maximal subgroup of $G$. Then $\Phi(G) \subseteq M$, whence $N' \subseteq M$, implying $M/N'$ is maximal in $G/N'$. Since $G/N'$ is nilpotent it follows that $M/N' \unlhd G/N'$ and by the correspondence theorem this is equivalent to $M \unlhd G$. So all maximal subgroups are normal, and $G$ is nilpotent.

!Proof of Lemma 1 If a group is nilpotent and let $M$ be a maximal subgroup. Then $M$ is normal and hence $G/M$ has no non-trival subgroups. This can only be if $M$ has prime index in $G$, whence $G/M$ is abelian and so $G' \subseteq M$. Since $\Phi(G)$ is the intersection of all maximal subgroup, we get $G' \subseteq \Phi(G)$. Conversely, if $G' \subseteq \Phi(G)$, then for any maximal subgroup $M$ we have $G' \subseteq \Phi(G) \subseteq M$, so $G' \subseteq M$ and $M$ is normal. 
Proof of Lemma 2 Observe that $\Phi(N)$ is characteristic in $N$ and $N$ is normal in $G$, so $\Phi(N)$ is normal in $G$. Hence for a maximal subgroup $M$ of $G$, $M\Phi(N)$ is a subgroup containing $M$. Assume $G=M\Phi(N)$. By Dedekind Lemma, we get $N=N \cap M\Phi(N)=(N \cap M)\Phi(N)=N \cap M \subseteq M$. Hence $\Phi(N) \subseteq N \subseteq M$ and it follows that $G=M\Phi(N)=M$ a contradiction. We conclude that $\Phi(N) \subseteq M$ for any maximal $M$, hence $\Phi(N) \subseteq \Phi(G)$ as required.

