Vector derivative of $f(x)=(A+x \mathbb{1}^T)^{-1} b$

Question

I am trying to find a vector derivative of $f(x)=(A+x \mathbb{1}^T)^{-1} b$ where $A \in \mathbb{R}^{n \times n}$ and $b \in \mathbb{R}^n$. Assume that all vectors a column. Also $\mathbb{1}$ is a vector of all ones.

I was trying to use the properties of vector derivatives that can be found here.
However, I could find a property I need.

I think it should be some thing of the form \begin{align} (A+x \mathbb{1}^T)^{-1} b \mathbb{1}^T (A+x \mathbb{1}^T)^{-1} \end{align}

Answer 1

Let's consider a more general problem by replacing $1$ with any constant column vector $c$. Then$$0=\partial_ib_j=\partial_i[(A_{jk}+x_jc_k)f_k]=(A_{jk}+x_jc_k)\partial_if_k+\delta_{ij}c_kf_k.$$The matrix $\partial f$ of $_{ik}$ index $\partial_if_k$ satisfies$$0=\partial f(A^T+cx^T)+(c\cdot f)I\implies\partial f=-(c\cdot f)(A^T+cx^T)^{-1}=-(A+xc^T)^{-1}bc^T(A^T+cx^T)^{-1}.$$

Answer 2

$ \def\a{\alpha}\def\b{\beta}\def\g{\gamma}\def\t{\theta} \def\l{\lambda}\def\s{\sigma}\def\e{\varepsilon} \def\n{\nabla}\def\o{{\tt1}}\def\p{\partial} \def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal G}} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\diag#1{\operatorname{diag}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \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}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} $For typing convenience, define the matrix variable and its various differentials $$\eqalign{ &M = \LR{A+x\o^T} \\ &dM = dx\,\o^T \\ &dM^{-1} = -M^{-1}\,dM\,M^{-1} \\ }$$ Write the function in terms of this new variable, then calculate its differential and gradient $$\eqalign{ f &= {M^{-1}}b \\ df &= -M^{-1}(dM)\;\c{M^{-1}b} \\ &= -M^{-1}\LR{dx\,\o^T}\c{f} \\ &= -\LR{\o^Tf}M^{-1}\,dx \\ \grad fx &= -\LR{f^T\o}M^{-1} \\ \\ }$$