Derivative of a function involving diagonal matrix Let $A$ be a $n\times n$ matrix, $\text{diag}(x)$ is the diagonal matrix with $x$ on the diagonal. How can I find $dF(x)$ for $F(x) = \text{diag}(x)Ax$? 
Thank you very much in advance!
 A: $
\def\d#1{\operatorname{Diag}\left(#1\right)}
$If you use the Hadamard (aka elementwise) product, then you can get rid of the diag operation, which will make it easier to find the differential and jacobian of the function.
$$\eqalign{
 f &= x\circ(Ax) \\
  &= \d{Ax}\,x 
  \;=\; \d{x}\,Ax \\
df &= \d{Ax}\,dx + \d{x}\,A\,dx \\
   &= \Big(\d{Ax} + \d{x}\,A\Big)\,dx \\
\frac{\partial f}{\partial x}
   &= \d{Ax} + \d{x}\,A \\\\
}$$
Special properties of the Hadamard product were used in several steps, i.e.
$$\eqalign{
x\circ y & = y\circ x \\
 &= \d{x}\,y \\
 &= \d{y}\,x \\
}$$
A: I assume that $x = (x_1, \dots, x_n)^T \in \mathbb{R}^n$ is a column vector so $F \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$. The differential $dF(x)$ must satisfy
$$ \lim_{h \to 0} \frac{F(x + h) - F(x) - dF(x)h}{\| h\|} = 0. $$
We have
$$ F(x + h) - F(x) = \operatorname{diag}(x + h)A(x + h) - \operatorname{diag}(x)Ax =  \\
(\operatorname{diag}(x) + \operatorname{diag}(h))(Ax + Ah) - \operatorname{diag}(x)Ax = \\
\operatorname{diag}(x)Ah + \operatorname{diag}(h)Ax + \operatorname{diag}(h)Ah.$$
The term $\operatorname{diag}(h)Ah$ is quadratic in $h$ so we can guess that $dF(x)h = \operatorname{diag}(x)Ah + \operatorname{diag}(h)Ax$ and prove it:
$$ \frac{\| F(x+h) - F(x) - \operatorname{diag}(x)Ah + \operatorname{diag}(h)Ax \|}{\| h \|} = \frac{\| \operatorname{diag}(h)Ah \|}{\| h \|} \leq \frac{ \| \operatorname{diag}(h) \| \| A \| \| h \|}{\| h \|} = \\
\| \operatorname{diag}(h) \| \| A \| \xrightarrow[h \to 0]{} 0.$$
