Consider $u (x) :\Omega\rightarrow\mathbb R^2$, $A(x):\Omega\rightarrow \mathbb R^{2\times 2} , x\in \Omega\subset\mathbb R^n$ and the following function $$f(x) = \exp(-u^T\cdot A^{-1}\cdot u )$$ I am interested in computing the Jacobian of $f,\, f_x$. I need to express this in terms of $u,\,u_x,\,A$ and $A_x$. I came up with the following:
$$ f_x = \frac{\mathrm df}{\mathrm dx}=f_AA_x+f_uu_x\\ =\left(-\left(u^T\cdot\frac{\mathrm dA^{-1}}{\mathrm dA}\cdot u\right)A_x -2(A\cdot u) \ u_x\right)\exp(-u^T\cdot A^{-1}\cdot u ) $$
But I have problems with computing the derivative of the inverse $A$ w.r.t. $A$. Also, intuitively I think that I've made some inner products and such.
Can anyone give me a push in the good direction?
Define the scalar variable $$\eqalign{\phi&=A^{-1}:uu^T\cr&=uu^T:A^{-1}}$$ where the colon is a handy product notation for the trace, i.e. $A:B={\rm tr}(A^TB)$
Write the function of interest in terms of this new variable, then find its differential. $$\eqalign{ f &= e^{-\phi} \cr \log f &= -\phi \cr \frac{df}{f} &= -d\phi \cr &= -\big(A^{-1}:d(uu^T)+uu^T:dA^{-1}\big) \cr &= -\big(A^{-1}:(du\,u^T+u\,du^T)-uu^T:A^{-1}\,dA\,A^{-1}\big) \cr &= A^{-T}uu^TA^{-T}:dA-u^T(A^{-1}+A^{-T})\,du \cr }$$ Since you have $(u_x,A_x)$ available, substitute them to obtain $$\eqalign{\frac{f_x}{f} &=(A^{-T}uu^TA^{-T}):A_x -u^T(A^{-1}+A^{-T})u_x \cr}$$ If $A$ is symmetric, you can simplify this result a bit. $$\eqalign{\frac{f_x}{f} &= (A^{-1}uu^TA^{-1}):A_x - 2u^TA^{-1}u_x \cr}$$ Since $A_x$ is a third-order tensor, it's tricky to work with. So instead of interpreting the colon as a trace function, it's better to use an explicit summation.
Let $v$ denote a vector, $M$ a matrix, and $T$ a tensor. $$v=M:T \implies v_k = \sum_{ij}M_{ij}{T}_{ijk}$$