I'm trying take the derivative of a symmetrix matrix $\mathbf{C}$ with respect to itself.
$$
\begin{equation}
\frac{\partial \mathbf{C}^{-1}}{\partial \mathbf{C}}
\end{equation}
$$
Using the indicial notation, above equation can be rewritten as follows
$$
\begin{equation}
\frac{\partial C_{ij}^{-1}}{\partial C_{kl}}
\end{equation}
$$
At first I've used the following formula,
$$
\begin{equation}
\frac{\partial C_{ij}^{-1}}{\partial C_{kl}} = -C^{-1}_{ik}C^{-1}_{lj}
\end{equation}
$$
But I quickly realized that we've lost the symmetry of the problem now.
I read The Matrix Cookbook and the other posts about the same problem but unfortunately, I couldn't understand the things they've done.
For example in this article, at Eq.(100) authors have used the property below when taking the derivative of Eq.(99)
$$
\begin{equation}
\frac{\partial \mathbf{C}^{-1}}{\partial \mathbf{C}} = -\mathbf{C}^{-1} \boxtimes \mathbf{C}^{-T} \mathbf{I}_s
\end{equation}
$$
Where $\boxtimes$ is the square product, $\mathbf{I}_s$ is the symmetric fourth-order identity tensor and they are defined as follows
$$
\begin{align}
(\mathbf{A} \boxtimes \mathbf{B})_{ijkl} &= \mathbf{A}_{ik}\mathbf{B}_{jl} \\
(\mathbf{I}_s)_{ijkl} &= \frac{1}{2}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})
\end{align}
$$
I couldn't understand how did they achieve this result and how can I derive it myself.
Best Answer
$ \def\p{\partial}\def\o{{\tt1}} \def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal G}} \def\C{C^{-1}}\def\Ct{C^{-T}} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} $Once you learn the technique, the problem can be solved very briefly $$\eqalign{ d\C &= -{\C\,dC\,\C} \\ &= -\LR{\C\E\Ct}:dC \\ \grad{\C}{C} &= -{\C\E\Ct} \\ }$$ The details are as follows...
Introduce a fourth-order tensor $\E$ with components $$\eqalign{ \E_{ijk\ell} = \delta_{ik}\,\delta_{j\ell} = \begin{cases} \o\quad{\rm if}\; i=k\;\;{\rm and}\;\;j=\ell \\ 0\quad{\rm otherwise} \end{cases} \\ }$$ The most useful property of this tensor is its ability to rearrange matrix products $$\eqalign{ ABC &= \LR{A\E C^T}:B \;=\; \F:B \\ }$$ where juxtaposition implies a single-dot product and a colon $(:)$ denotes the double-dot product $$\eqalign{ &\F_{ijk\ell} = \sum_{p=1}^n\sum_{r=1}^n A_{i\c{p}}\E_{\c{p}jk\c{r}}\,C_{\c{r}\ell}^T \;=\; A_{ik}C_{\ell j} \\ &\LR{\F:B}_{ij} = \sum_{k=1}^n\sum_{\ell=1}^n \F_{ij\c{k\ell}}\,B_{\c{k\ell}} = \sum_{k=1}^n\sum_{\ell=1}^n A_{i\c{k}}B_{\c{k\ell}}C_{\c{\ell}j} \\ }$$ Start with the differential of the matrix inverse identity $$\eqalign{ &I = \C C \\ &dI = \c{d\C}C + \C dC \;\doteq\; 0 \\ &\c{d\C} = -\C\,dC\,\C \\ }$$ Then use $\E$ to rearrange the terms and recover the gradient $$\eqalign{ d\C &= -{\C\E\Ct}:dC \\ \grad{\C}{C} &= -{\C\E\Ct} \\ }$$ Or in component notation $$\eqalign{ \grad{\C_{ij}}{C_{k\ell}} &= -\sum_{p=1}^n\sum_{r=1}^n \C_{i\c{p}}\E_{\c{p}jk\c{r}}\C_{\ell\c{r}} \;=\; -\C_{ik}\C_{\ell j} \\ }$$
Update
The comments have become a rehash of the old "symmetric gradient" debate.
On the other hand, if a small set of scalar parameters are used to construct a tensor quantity, then the derivative of the tensor components with respect to one of those scalar parameters can exhibit any number of interesting symmetries. A large part of Continuum Mechanics is devoted to studying the implications of such symmetries.
But that's a different problem than calculating the derivative of one tensor component with respect to another tensor component. But many people (even professors and famous authors) often conflate these two problems.