How to prove this matrix differential for Born-Infeld theory? Consider the Born-Infeld Lagrangian, page 30 of Born-Infeld Action and Its
Applications  by Cong Wang.
$L_{BI} = \sqrt{\det (1+ F)}$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. I am trying to derive the EOM as done here. However I coudn't follow several steps. I am uncomfortable with the author treating $F_{\mu\nu}$ as a number without indices. If I follow the first few lines, it was shown that $L_{BI}$ becomes ${\det (1- F^2)}^{\frac{1}{4}}$ where $({F^2})^{\alpha\beta} = F^{\alpha\sigma}F^{\beta}_{\sigma}$
Following through the next steps,
$$\delta L_{BI} = \delta \exp(\frac{1}{4}\mathrm{tr} \ln(1-F^2))
=-\frac{1}{2} \exp(\frac{1}{4}\mathrm{tr} \ln(1-F^2)) (\frac{F}{1-F^2})^{\mu \nu} \delta F_{\nu \mu}$$
$$\delta \mathrm{tr} \ln(1-F^2) = (\frac{F}{1-F^2})^{\mu \nu} \delta F_{\nu \mu}$$
I am trying to understand this. Is the trace taken before taking the derivative? I would appreciate any help in proving this starting from $ \delta \mathrm{tr} \ln(1-(F^2)^{\alpha \beta})$. What matrix identies are to used?
 A: For any matrix $A$:
$$\delta \mathrm{tr}\, A = \mathrm{tr}\, \delta A$$
since $\mathrm{tr}$ is just a linear combination of matrix elements of its argument, and $\delta$ is linear. Writing the indices explicitly ($\delta_{ij}$ is Kronecker's delta):
$$ \delta \, \mathrm{tr} A = \delta A_{ii} = \delta (\delta_{ij} A_{ij}) = \delta_{ij} \delta A_{ij} = \delta_{ij} (\delta A)_{ij} = \mathrm{tr} \, \delta A$$
Now let's take $A= \log (1-M^2) $ and go on:
$$
\mathrm{tr}\,\delta \log (1-M^2) =\mathrm{tr}\,\left[ -(1-M^2)^{-1} \delta (M^2) \right]= -2 ({(1-M^2)^{-1}})_{ij} M_{jk} \delta M_{ki}
$$
now writing
$$
({(1-M^2)^{-1}})_{ij} M_{jk} = \left(\frac{M}{1-M^2}\right)_{ik}
$$
you have your result for any matrix $M$.
In the present case, things are slightly different because you have to treat Lorentz indices properly,  and trace are computed by contracting with the inverse metric $\eta^{\mu\nu}$, so for instance
$$
\mathrm{tr}\,\left[(1-M^2)^{-1} \delta (M^2) \right] = 
((1-M^2)^{-1})^{\alpha\beta} \delta (M^2)_{\beta\alpha},
$$
but the key argument is the one expressed before.
