If G/N is a cyclic group then, [G,G] =[G,N] We want to show that, if $N$ is a normal subgroup of $G$ (finite p-group) such that $G/N$ is cyclic, then $[G,G]=[G,N]$.
I have done following. Since, $G/N$ is cyclic, $G/N = ⟨xN⟩$. Also, we know if $G/N$ is abelian then $[G,G] < N$. Hence if we show $G/[G,N]$ is abelian then we are done. Now showing $G/[G,N]$ is abelian boils down to showing $[g_1, g_2]$ belongs to $[G,N]$. But I can't manipulate to prove it. 
 A: In general the following holds.
Proposition Let $G=HN$ for a subgroup $H$ and normal subgroup $N$, then $G'=H'[G,N]$.
 Observe that the answer to your question follows: if $G/N$ is cyclic, then $G/N=\langle \overline{x} \rangle$ for some $x \in G$, where $\bar{.}$ denotes modding out by $N$. It follows that $G=\langle x \rangle N$. Take $H=\langle x \rangle$ in the proposition and note that $H'=\{1\}$.You do not even need the $p$-group condition.
Proof of the Proposition: it is straightforward to see that $[G,N] \unlhd G$ and that $\overline{N} \subseteq Z(\overline{G})$, where for the moment  $\bar{.}$ denotes modding out by $[G,N]$. Since $G=HN$, we get $\overline{G}=\overline{H}\overline{N}$ and using $\overline{N}$ being central in $\overline{G}$ it follows that $\overline{G}'=\overline{H}'$. This is equivalent to $\overline{G'}=\overline{H'}$. We conclude that if $g \in G'$, then $g=hn$ for some $h \in H'$ and $n \in [G,N]$, that is $G' \subseteq H'[G,N]$ and the reverse inclusion is trivial $\square$.

Note The proposition can be further generalized to varieties of groups. If $\mathcal{V}$ is a variety of groups and let $V^*(G)$, respectively $V(G)$ denote the marginal and verbal subgroup of $G$ w.r.t. to the variety $\mathcal{V}$.
If $G=HN$, with $N$ normal, then $V(G)=V(H)[NV^*G]$. For a proof and further details see Theorem(2.4) in my paper. Example: if $\mathcal{V}=\mathcal{N}_c$, the variety of nilpotent groups of class $c$, then $\gamma_{c+1}(G)=\gamma_{c+1}(H)[N,_cG]$.
A: I also figured out another easy way to solve the question. Please let me know if this works also. We know by the definition of the commutator subgroup that $[G,N]\subseteq [G,G]$. Hence we will show that $[G,G]\subseteq [G,N]$. We will show that the generators of $[G,G]$ lie in $[G,N]$ and this will prove the result. Now, since $G/N$ is cyclic, $\Rightarrow G/N = \langle xN \rangle$ for some $x\in G$. This also means that for any $gN\in G/N , gN = x^{i}N $ for some integer $i$. $\Rightarrow g = x^{i}n, $ for some $n\in N$
Now, for any generator $[g_1,g_2]\in [G,G] , [g_1, g_2] = [x^{i_1}n_1, x^{i_2}n_2].$ Now, we will use the following two commutator formulas:
(1) $[xy,z] = [x,z][x,z,y][y,z]$
(2) $[x,yz] = [x,z][x,y][x,y,z]$
Then, $[g_1,g_2] =[x^{i_1}n_1, x^{i_2}n_2] = [x^{i_1},x^{i_2}n_2][x^{i_1},x^{i_2}n_2,n_1][n_1,x^{i_2}n_2] $ using (1). Now the second and third commutators are in $[G,N]$. Hence need to show that the first one is also in $[G,N]$ We will use (2) now.
$[x^{i_1},x^{i_2}n_2] = [x^{i_1},n_2][x^{i_1},x^{i_2}][x^{i_1},x^{i_2},n_2] = [x^{i_1},n_2],$ as last two commutators become $1$. Hence we have finally, 
$[g_1,g_2] =  [x^{i_1},n_2][x^{i_1},x^{i_2}n_2,n_1][n_1,x^{i_2}n_2].$ This has all commutators in $[G,N]$. Hence $[G,G] \subseteq [G,N]$ and we are done. 
