Conserved currents for Lagrangian given by a trace Let the Lagrange density $\mathcal{L}$ be given by
$\mathcal{L}=\mathrm{Tr}\left(\partial_\mu U^\dagger \partial^\mu U\right)$,
where $U=U(x)\in U(N)$. Suppose there are two matrices $A,B\in SU(N)$ and consider transformation $U(x)\rightarrow A^\dagger U(x)B$. One can easily check that $\mathcal{L}$ is invariant under such transformation, hence by Nother's theorem there exist conserved currents $j_{A,B}$ depended of choice of matrices $A,B$. My question is how to derive these currents ? I know the general formula given in the proof of Noether's theorem but I don't how to treat derivatives $\frac{\partial \mathcal{L}}{\partial (\partial_\mu U(x))}$ in this case. Moreover, we need to obtain $\delta U(x)$. I think the last one can be obtained using representation of $A,B$, which are near to unity, by the exponent of generators of appropriate Lie algebra. 
 A: Seeing that it is tagged as a homework exercise, I'm probably not supposed to solve the problem, but I can give some hints as to how I would tackle the problem. Assuming one wants to use the standard approach to derive the Noether current, I would suggest that one does not overthink the problem and simply treat the $U$'s as (matrix-valued) functions. One can then apply the derivatives as functional derivatives in the usual way (but I'll use the normal notation for derivatives). One would need a rule for applying the functional derivatives. Making the indices of the matrices explicit, one would have
$$ \frac{\partial [U(x)]_{ab} }{\partial[U(y)]_{cd}} = \delta(x-y)\delta_{a}^{c}\delta_{b}^{d} . $$
You can now generalize this to the case where $U$ is replaced by $\partial_{\mu} U$.
Another important thing to remember is that the derivative of the adjoint is not zero.
$$ \frac{\partial U^{\dagger} }{\partial U}\neq 0 . $$
For this one can use
$$ U^{\dagger} = U^{\dagger}UU^{\dagger} . $$
If you still get stuck, let me know, then I can perhaps expand on some of these points.
