The derived subgroup $[G,G]$ is a fully invariant subgroup because if $f$ is an endomorphism of $G$ , then for each word of commutators $[a_1,b_1][a_2,b_2]\cdots[a_m,b_m]$ , we have $$f([a_1,b_1][a_2,b_2]\cdots[a_m,b_m])=[fa_1,fb_1][fa_2,fb_2]\cdots[fa_m,fb_m]\in [G,G]$$ i.e. the homomorphic image of a word of commutators is a word of commutators.