Vector calculus: Compute Jacobian $D_x (A(x)^{-1} v(x))$ I wish to compute
$$
D_x \left( A(x)^{-1} v(x) \right)
$$
where $x$ is a vector, $A$ is a matrix-valued function such that $A(x)$ is always invertible, and $v(x)$ is a vector-valued function.
Looking here, I know that
$$
\left( D_x \left( A(x)^{-1} v(x) \right)\right)_{ij}=\sum_k \left( \frac{\partial A(x)_{ik}^{-1}}{\partial x_{j}} v(x)_k  + A(x)^{-1}_{ik} \frac{\partial v(x)_k}{\partial x_j}\right). 
$$
The second term in the sum is easy, and is just $(A(x)^{-1} D_x v(x))_{ij}$ where $D_x v(x)$ is the Jacobian of $v$. For the first term, I do not know how to compute
$$
\frac{\partial A(x)_{ik}^{-1}}{\partial x_{j}}.
$$
I tried to use the result given here, which says that if $x$ is a scalar variable then
$$
\frac{\partial A(x)^{-1}}{\partial x}=-A(x)^{-1} \frac{\partial A(x)}{\partial x}A(x)^{-1},
$$
but I cannot extend it to vector variable $x$ and to the three-dimensional tensor
$$
\left( D_xA(x)^{-1}\right)_{ijk} = \frac{\partial A(x)_{ik}^{-1}}{\partial x_{j}}.
$$
 A: To try to reduce clutter, I will use the following notation for the derivative:
$$
  D^h[f(x)]
$$
where $x$ is implicitly being differentiated, $h$ is the point at which the linear map $D[f(x)]$ is evaluated. We will also use the notation
$$
  D^h[f]_y
$$
To denote the derivative of $f$ at the point $y$, with the linear map $D[f]_y$ evaluated at $h$. We have the equality
$$
  D^h[f(x)] = D^h[f]_x.
$$
Additionally, an expression like
$$
  \dot D[f(\dot x)g(x)]
$$
means that only $\dot x$ is being differentiated, and the undotted $x$ is held constant; in a more verbose notation
$$
  \dot D[f(\dot x)g(x)] = \Bigl[D_y[f(y)g(x)]\Bigr]_{y=x}.
$$

Applications of the chain rule give
$$\begin{aligned}
D^h[A(x)^{-1}v(x)]
  &= \dot D^h[A(\dot x)^{-1}v(x)] + \dot D^h[A(x)^{-1}v(\dot x)]
\\
  &= D^h[A(x)^{-1}]v(x) + A(x)^{-1}D^h[v(x)].
\end{aligned}$$
Now let $I(X) = X^{-1}$ be the matrix inversion map so that $A(x)^{-1} = (I\circ A)(x)$. We can then use the chain rule to write
$$
  D[I\circ A]_x = D[I]_{A(x)}\circ D[A]_x,
$$
and we need only determine $D[I]$. From the defining equation of $I$,
$$\begin{aligned}
  I(X)X = 1
  &\implies \dot D^H[I(\dot X)X] + \dot D^H[I(X)\dot X] = 0
\\
  &\implies D^H[I(X)]X + I(X)H = 0
\\
  &\implies D^H[I(X)] = -X^{-1}HX^{-1}.
\end{aligned}$$
Thus
$$
  D^h[A(x)^{-1}] = D^h[I\circ A]_x = -A(x)^{-1}D^h[A(x)]A(x)^{-1}.
$$
Finally, we have
$$
  D^h[A(x)^{-1}v(x)] = -A(x)^{-1}D^h[A(x)]A(x)^{-1}v(x) + A(x)^{-1}D^h[v(x)].
$$
A: Actually, you can express your derivative in a more compact form.
Since
$v(x)=A(x)(A(x)^{-1}v(x))$ (Eq. 1)
differentiating (Eq.1) w.r.t. $x$, gives:
$\frac{d}{dx}(v(x))=(\frac{d}{dx}A(x))A(x)^{-1}v(x)+A(x)\frac{d}{dx}(A(x)^{-1}v(x))$ (Eq. 2)
Thus, solving (Eq. 2) for $d(A(x)^{-1}v(x))/dx$ yields:
$\frac{d}{dx}(A(x)^{-1}v(x))=A(x)^{-1}[\frac{d}{dx}v(x)-(\frac{d}{dx}A(x))(A(x)^{-1}v(x))]
$ (Eq.3)
Notice that Eq. (3)  expresses the derivative of $A^{-1}v(x)$ as only function of the derivatives of $A(x),v(x)$ that are known in your problem. Also, $dv(x)/dx$ is an $n\times n$  matrix and $dA(x)/dx$ is an $n\times n\times n$ tensor. If you are not familiar about matrix derivatives with respect to a vector, have a look on some Matrix Calculus source, e.g. Click here
Example:
Let $v(x)=\begin{bmatrix}
v_1(x)\\
v_2(x)
\end{bmatrix}$ and $x=[x_1\,x_2\,x_3]$. Then:
$\frac{dv(x)}{dx}=\begin{bmatrix}
\frac{\partial v_1(x)}{\partial x_1}&\frac{\partial v_1(x)}{\partial x_2}&\frac{\partial v_1(x)}{\partial x_3}\\
\frac{\partial v_2(x)}{\partial x_1}&\frac{\partial v_2(x)}{\partial x_2}&\frac{\partial v_2(x)}{\partial x_3}
\end{bmatrix}$
