Solution
= Solution
The first term $gA_\mu g^{-1}$ lies in $\mathfrak g$ because the <adjoint action> preserves the Lie algebra. For the second, fix $x$ and consider the group curve $s\mapsto g(x+s e_\mu)g(x)^{-1}$ through the identity. Its tangent at zero is $(\partial_\mu g)g^{-1}$, so this is also in $\mathfrak g$. Since a <Lie algebra> is a vector space, $A'_\mu\in\mathfrak g$.