How to show that the commutator subgroup is a normal subgroup
Denote the commutator of $a$ and $b$ by $a^{-1}b^{-1}ab = [a,b]$.
If $u$ is an element from the commutator subgroup, then $g^{-1}ug = u(u^{-1}g^{-1}ug) = u[u, g]$ .
Another approach: the commutator subgroup is defined to be the subgroup generated by the commutators, so every element of the commutator subgroup is of the form $$[a_1, b_1][a_2,b_2]\ldots[a_n, b_n].$$ It is enough to show that $g^{-1}[a,b]g$ is always in the commutator subgroup, because then
$$g^{-1}[a_1, b_1][a_2,b_2]\ldots[a_n, b_n]g = (g^{-1}[a_1, b_1]g)(g^{-1}[a_2,b_2]g)(g^{-1}\ldots g)(g^{-1}[a_n, b_n]g)$$
is a product of elements from the commutator subgroup. When $\phi$ is any homomorphism, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Since for any $g \in G$ the map $\phi$ defined by $\phi(x) = g^{-1}xg$ is a homomorphism, the result follows.
If $c \in G'$ and $g \in G$, then also $[g, c] = gc g^{-1} c^{-1} \in G'$. Because G' is closed under products, we would also have $(gc g^{-1} c^{-1})c \in G'$. But $(gc g^{-1} c^{-1})c = gcg^{-1} \in G'$, so by definition $G'$ is a normal subgroup of $G$.