Derivative of real antisymmetric matrix in mathematica

Is it possible to find the derivative of components of a real antisymmetric matrix using index notation? Eg: I have a very large real antisymmetric matrix. Then from Matrix Cookbook, we know the formula that: $$\frac{dx_{k\ell}}{dx_{\beta\alpha}}=\delta_{k\beta}\delta_{l\alpha}-\delta_{k\alpha}\delta_{l\beta}$$ This formula is a general one and can be applied to any indices. So is there a way to define this in Mathematica for any general indices? That means even if I use, $$\frac{dx_{ab}}{dx_{cd}}$$,it should return the answer without any other definition. Is this possible? Please help as I am new to Mathematica. Some other equations that can be obtained from the above relation of differentiation real antisymmetric matrices are given as: $$\frac{d}{dx_{\beta\alpha}}\left(x_{\upsilon l}x_{k\mu}\right)=\left(\delta_{\upsilon\beta}\delta_{l\alpha}-\delta_{\upsilon\alpha}\delta_{l\beta}\right)x_{k\mu}+x_{\upsilon l}\left(\delta_{k\beta}\delta_{\mu\alpha}-\delta_{k\alpha}\delta_{\mu\beta}\right)$$ $$\frac{d}{dx_{\beta\alpha}}\left(x_{k\alpha}x_{\beta l}\right)=-2\left(D-1\right)x_{lk}=2\left(D-1\right)x_{kl}$$ $$\frac{d}{dx_{\beta\alpha}}x_{j\alpha}=\delta_{j\beta}\delta_{\alpha\alpha}-\delta_{j\alpha}\delta_{\alpha\beta}=\left(D-1\right)\delta_{j\beta}$$ But my issue is that this is not the only indices I need to use. I need to find out the same with other indices. So is there is a way that Mathematica can find it automatically. Also in the below expression, I have implemented the product rule. $$\frac{d}{dx_{\beta\alpha}}\left(x_{\upsilon l}x_{k\mu}\right)=\left(\delta_{\upsilon\beta}\delta_{l\alpha}-\delta_{\upsilon\alpha}\delta_{l\beta}\right)x_{k\mu}+x_{\upsilon l}\left(\delta_{k\beta}\delta_{\mu\alpha}-\delta_{k\alpha}\delta_{\mu\beta}\right)$$ So if these things can be calculated by Mathematica, then I can check my calculations. D is the dimension of the real antisymmetric matrices. $$ \sum_{ijkl}f_{ijkl}=-\sum_{ijkl}f_{jikl} $$ The above tensor f is defined in such a way that we can exchange it's indices such a way that one exchange of neighboring indices results in a negative sign. Two nearby exchanges result in no change in sign. $$ \sum_{ijkl}f_{ijkl}x_{k\alpha}^{-}\frac{dx_{ij}^{+}}{dx_{\beta\alpha}}x_{\beta\ell}^{+} $$ $$= \sum_{ijkl}f_{ijkl}x_{k\alpha}^{-}\left(\delta_{i\beta}\delta_{j\alpha}-\delta_{i\alpha}\delta_{j\beta}\right)x_{\beta\ell}^{+}$$ $$=\sum_{ijkl}f_{ijkl} x_{k\alpha}^{-}x_{\beta\ell}^{+}\delta_{i\beta}\delta_{j\alpha}-\sum_{ijkl}f_{ijkl}x_{k\alpha}^{-}x_{\beta\ell}^{+}\delta_{i\alpha}\delta_{j\beta}$$ $$=\sum_{ijkl}f_{ijkl} x_{kj}^{-}x_{i\ell}^{+}-\sum_{ijkl}f_{ijkl}x_{ki}^{-}x_{j\ell}^{+}$$ $$= \sum_{ijkl}f_{ijkl}x_{kj}^{-}x_{i\ell}^{+}-\sum_{jikl}f_{jikl}x_{kj}^{-}x_{i\ell}^{+}$$ $$= \sum_{ijkl}f_{ijkl}x_{kj}^{-}x_{i\ell}^{+}+\sum_{ijkl}f_{jikl}x_{kj}^{-}x_{i\ell}^{+}$$ $$= 2\sum_{ijkl}f_{ijkl}x_{kj}^{-}x_{i\ell}^{+} $$ Utilizing the antisymmetric property of x matrices we can rewrite the above expression as: $$ \sum_{ijkl}f_{ijkl}x_{k\alpha}^{-}\frac{dx_{ij}^{+}}{dx_{\beta\alpha}}x_{\beta\ell}^{+}=-2\sum_{ijkl}f_{ijkl}x_{jk}^{+}x_{i\ell}^{+} $$ We have $$ x_{kj}^{-}=x_{kj}-i\delta_{kj}=-x_{jk}-i\delta_{kj}=-x_{jk}^{+} $$ When I try to evaluate the following code:

myD[x[v, l] - x[β, α], x[p, u]] 

with the following before,

myD[a_ - b_, c_] := myD[a, c] - myD[b, c] myD[a_ + b_, c_] := myD[a, c] + myD[b, c] 

It's returning zero.