Derivative of $F(Ax)$ What is the identity for $$ \frac{\partial \mathbf{F}(\mathbf{A}\mathbf{x})}{\partial \mathbf{x}} = ?$$
If $\mathbf{A} \in \mathbb{R}_{mn}$, $\mathbf{x} \in \mathbb{R}_n$, and $\mathbf{F}: \mathbb{R}^m \rightarrow \mathbb{R}^m $ , where $\mathbf{F}(\mathbf{x}) = [f(x_1) \; f(x_2) \; ... \; f(x_m)]^T $ and $ f: \mathbb{R} \rightarrow \mathbb{R} $
I didn't find it here or there.
 A: Just use the chain rule.  Linear operators are essentially vector fields; they're just linear.  This viewpoint makes the application of multivariable calculus very straightforward.
Let $x' = Ax$.  Let $\nabla$ be the usual vector derivative (that is, $\partial_x$) and $\nabla'$ be that corresponding to the dimension of $x'$.  Let $a$ be a vector, and the chain rule tells us that
$$a \cdot \nabla (F \circ A)(x) = [a \cdot \nabla A(x)] \cdot \nabla' F(x')$$
$A$ is a linear function, so $a \cdot \nabla A(x) = A(a)$, so the result is
$$[A(a) \cdot \nabla'] F(x')$$
You can then evaluate the components by plugging in basis vectors for $a$.  In index notation, this yields
$$\sum_{j'} {A_i}^{j'} \partial_{j'} F_{k'}$$
This has two free indices, so the result can be interpreted as a matrix of partial derivatives of the component functions.
A: $\def\D{\operatorname{Diag}}\def\p{{\partial}}\def\grad#1#2{\frac{\p #1}{\p #2}}\def\hess#1#2#3{\frac{\p^2 #1}{\p #2\,\p #3^T}}$For
typing convenience, introduce the vector variable
$$\eqalign{
y &= Ax \quad\implies\quad dy = A\,dx \\
}$$
Denote the scalar function and its derivative as $\big(f,f'\big)\,$ and
denote their element-wise application on the new vector variable by the vectors
$$\eqalign{f &= f(y) \qquad f' &= f'(y)}$$
It will also be convenient to use an uppercase letter to denote the matrix
$$F' = \D(f')$$
Write the differential of the function using the elementwise/Hadamard product,
replace the Hadamard product with the diagonal matrix, then perform a change of variables from $y\to x\,$ to find the desired gradient.
$$\eqalign{
df &= f'\odot dy \;=\; F'dy \;=\; F'A\,dx \\
\grad{f}{x} &= F'A \\
}$$
In your specific example
$$\eqalign{
f &= \sin(Ax) \qquad f' = \cos(Ax) \\
\grad{f}{x} &= \D\left(\cos(Ax)\right)\,A \\
}$$
