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I would like to calculate the second order derivative of a function with respect to a matrix coefficient symbolically. Below is the first order derivative of S with respect to the coefficient of x. For getting the answer what I have done is I have expressed the S as the function of only coefficients.

myD[S , x_[n_, s_]] := D[S[x[n, s]], x[n, s]];
   

Here S is a function of the big matrix 'x'. But as I said before I need to find the derivative of S with respect to the coefficient of x.

myD[myD[S, x_[n_, s_]], x_[t_, r_]] := 
  D[S[x[n, s]], {x[n, s], 2}];

Above code is what I have written for second derivative and of course it is not correct as what I need is something as below equation.

Is there a best way to define that? $$\sum_{\nu<\mu}\sum_{\beta<\alpha}x_{i\alpha}x_{\beta j}x_{k\mu}x_{\nu\ell}\frac{\partial^{2}S}{\partial x_{\nu\mu}x_{\beta\alpha}}$$ This is how it participate. Definition: $$\frac{\partial^{2}S}{\partial x_{\nu\mu}x_{\beta\alpha}}=\frac{\partial}{\partial x_{\nu\mu}}\left(\frac{\partial S}{\partial x_{\beta\alpha}}\right)$$

If we have, $$K=x_{l\alpha}x_{\beta k}x_{i\nu}x_{\mu j}$$

$$\frac{\partial SK_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\upsilon\mu}}=S\frac{\partial K_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\upsilon\mu}}+K\frac{\partial S}{dx_{\upsilon\mu}}$$

$$\frac{\partial^{2}SK_{\left(\alpha\beta\right),\left(\mu\upsilon\right)}}{dx_{\beta\alpha}x_{\upsilon\mu}} =\frac{\partial}{dx_{\beta\alpha}}\left(\frac{\partial SK_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\upsilon\mu}}\right) =\frac{\partial}{dx_{\beta\alpha}}\left(S\frac{\partial K_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\upsilon\mu}}+K\frac{\partial S}{dx_{\upsilon\mu}}\right) =\left(S\frac{\partial^{2}K_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\beta\alpha}x_{\upsilon\mu}}+\frac{\partial K_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\upsilon\mu}}\frac{\partial S}{dx_{\beta\alpha}}+K\frac{\partial^{2}S}{dx_{\beta\alpha}x_{\upsilon\mu}}+\frac{\partial K_{\left(\beta\alpha\right),\left(\upsilon\mu\right)}}{dx_{\beta\alpha}}\frac{\partial S}{dx_{\upsilon\mu}}\right) $$

provided D is the dimension of the matrix. Also, $$ \frac{\partial x_{lj}}{dx_{\beta\alpha}}=\delta_{j,\alpha}\delta_{l,\beta}-\delta_{j,\beta}\delta_{l,\alpha}$$

EDITED But when I put the rule of differentiation of real antisymmetric matrix as in the above equation by the following code, it is showing error.

myD[x_[k_, l_], 
   x_[v_, g_]] := \[Delta][k, v] \[Delta][l, g] - \[Delta][k, 
     g] \[Delta][l, v];
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    $\begingroup$ This really depends on how the derivative of S participates in subsequent calculation. Can you show a few examples? $\endgroup$
    – xzczd
    Jul 24, 2020 at 1:29
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    $\begingroup$ Interesting, I didn't notice this behavior before. It's because stuff inside Format is automatically evaluated (yeah, though := owns the attribute HoldAll). Luckily it's easy to resolve, just add a HoldPattern, see my update. $\endgroup$
    – xzczd
    Jul 24, 2020 at 12:36
  • $\begingroup$ @xzczd No words☺️☺️Amazing $\endgroup$
    – Jasmine
    Jul 24, 2020 at 12:42

1 Answer 1

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If it's just for display, then I think the following is enough:

Clear[myD, x]
myD[myD[a_, b__], c__] := myD[a, b, c]

Format[HoldPattern@myD[a_, b__]] := TraditionalForm@HoldForm@D[a, b]

Format[x[a_, b_]] := Subscript[x, a, b]

myD[myD[S, x[ν, μ]], x[α, β]]

enter image description here


To automate the subsequent calculation, just define

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

and the following two lines that are already in your previous question:

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

Format[x[a_, b_]] := Subscript[x, a, b]

Check:

myD[myD[S K, x[β, α]], x[ν, μ]]

enter image description here

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