If a finite group $G$ is solvable, is $[G,G]$ nilpotent?

The symmetric group $S_4$ is solvable, but $[S_4, S_4] = A_4$ is not nilpotent.

As mentioned in another answer, this is unfortunately not true (another example is $GL_2(3)$ whose derived subgroup is $SL_2(3)$). One interesting thing to note is that if it was true, then this would give an easy proof of the conjecture that the Taketa inequality holds for all solvable groups.

Elaboration: The Taketa inequality is $\rm{dl}(G)\leq |\rm{cd}(G)|$ where $\rm{dl}(G)$ is the derived length of $G$ and $\rm{cd}(G) = \{\chi(1) | \chi\in \rm{Irr}(G)\}$ is the set of degrees of irreducible complex characters of $G$.

As mentioned, this inequality is conjectured to hold for all solvable groups, and the way it would follow if the derived subgroup was nilpotent is due to a theorem of Isaacs and Knutson, which states:

If $N$ is a normal nilpotent subgroup of $G$ then the derived length of $N$ is at most the number of degrees of irreducible complex characters of $G$ which do not have $N$ in their kernel.

Using the above theorem with $N = G'$ one immediately gets the conjecture.