# How to take the determinant of a rank(1,1) tensor?

I want to find the Jacobian matrix and its determinant of the generic infinitesimal transformation:

$$x'^\mu=x^\mu+\epsilon_\alpha\frac{\delta x^\mu}{\delta \epsilon_\alpha}$$ where $$\epsilon_\alpha$$ is small and might be dependant on $$x$$.

I think the Jacobian matrix is

$$\partial_\nu x'^\mu=\delta^\mu_\nu+\frac{\delta x^\mu}{\delta \epsilon_\alpha}\partial_\nu\epsilon_\alpha+\epsilon_\alpha\partial_\nu\frac{\delta x^\mu}{\delta \epsilon_\alpha}$$ where $$\partial_\nu=\frac{\partial}{\partial x^\nu}$$

where I suspect the last term to be zero since $$\partial_\nu\frac{\delta x^\mu}{\delta \epsilon_\alpha}=\frac{\delta \partial_\nu x^\mu}{\delta \epsilon_\alpha}=\frac{\delta \delta^\mu_\nu}{\delta \epsilon_\alpha}=0$$ in other words I suspect the derivative to commute with the functional derivative. So we get:

$$\partial_\nu x'^\mu=\delta^\mu_\nu+\frac{\delta x^\mu}{\delta \epsilon_\alpha}\partial_\nu\epsilon_\alpha$$.

Now I want to know the determinant of this expression but am not totally sure how to go about getting this properly.

The reason that I want this is to derive a general expression for the conserved current that I had previously not encountered:

$$j_\alpha^\mu=\Big[\frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi)}\partial_\nu \Phi-\delta^\mu_\nu\mathcal{L}\Big]\frac{\delta x^\nu}{\delta \epsilon_\alpha}-\frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi)}\frac{\delta F}{\delta \epsilon_\alpha}$$ where $$F=\Phi'(x')$$

This is encountered in the lecture notes of Joshua D. Qualls called "Lectures on Conformal Field Theory". The expression is given on p.35. There is an exercise on p.162 that is supposed to guide you through the derivation.