$\frac{\partial A^{-1}} {\partial x}$ = $-A^{-1} \frac{\partial A}{\partial x} A^{-1}$ If A(x) is invertible A(x):  $\Bbb R$ $\rightarrow$ $\Bbb R^{(n\times n)}$ is a map from $\Bbb R$ to $\Bbb R^{n\times n}$Show that $\frac{\partial A^{-1}} {\partial x}$ = $-A^{-1} \frac{\partial A}{\partial x} A^{-1}$
What I know is that since A is invertible. So, $A^{-1}(x)$: $\Bbb R^{(n\times n)}$ $\rightarrow$ $\Bbb R$. So, $[A^{-1}]$ is a n by n matrix, and its gradient is
$\frac{\partial A^{-1}}{\partial x} = $$
    \begin{bmatrix}
    \frac{\partial A^{-1}}{\partial x_{11}} & ... & \frac{\partial A^{-1}}{\partial x_{1n}} \\
    .\\.\\.\\
    \frac{\partial A^{-1}}{\partial x_{n1}} & ... & \frac{\partial A^{-1}}{\partial x_{nn}} \\
    \end{bmatrix}
$
So, we just need to show that  $A\frac{\partial A^{-1}} {\partial x}A$ = $- \frac{\partial A}{\partial x}$
However, I am stuck at what is $\frac{\partial A}{\partial x}$. A is a map from a 1 by 1 into an n by n matrix. I am confused that what is A? Please give me some advice, thanks!

The easiest way to prove the identity is to take the $\frac{\partial}{\partial x}$ derivative of the equation $A A^{-1}=1$. By the chain rule we get $$\frac{\partial A} {\partial x} A^{-1}+A \frac{\partial (A^{-1})}{\partial x}=0$$ which implies your statement.
