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is an normal real valued function. If you want you can write it componentwise as
$$f(x) = {1\over 2}\sum_j\sum_k p_{jk}x_jx_k + \sum_j q_jx_j + r$$
Now the first double sum contains the $x_jx_k$ term twice if 
and if 
it becomes an 
term, so the derivate with respect to 
becomes:
Which in matrix notation becomes
I simply would use the Gâteaux-Derivative. That derivative is the natural expansion of the 1D Derivative
to higher dimensions.
Since your function maps $f:ℝ^n→ℝ$ we need an arbitrary direction $δx∈ℝ^n$, and a small increment 
. Using that "
formulation the Gâteaux-Derivative for your function reads
\begin{align*}
d(\|Ax-b\|²;[x,δx]) = (\frac{d}{dε}\|A(x+εδx) - b\|²)\big|_{ε=0}
\end{align*}
First it is \begin{align*} \frac{d}{dε}\|A(x+εδx) - b\|² =& \frac{d}{dε}[(A(x+εδx) - b, A(x+εδx) - b)] \\ =&\frac{d}{dε}[\{(Ax, Ax)+ (Ax,Aεδx) + (Ax, -b)\} \\ &+ \{(Aεδx, Ax) + (Aεδx, Aεδx) + (Aεδx, -b)\} \\ &+ \{(-b, Ax) + (-b, Aεδx) + (-b, -b)\} ] \\ =¹&\frac{d}{dε}[\{\|Ax\|²+ \|b\|²+ 2(Ax, -b)\} \\ &+ ε\{2(Ax,Aδx) + 2(-b, Aδx)\} \\ &+ ε²\|Aδx\|² ]\\ =& \{2(Ax,Aδx) + 2(-b, Aδx)\} + 2ε\|Aδx\|². \end{align*} ¹Sorting by powers of ε.
Setting ε=0, yields \begin{align*} (\frac{d}{dε}\|A(x+εδx) - b\|²)\big|_{ε=0} &= 2(Ax,Aδx) + 2(-b, Aδx) \\ &= 2(Ax-b, Aδx)= (2A^\top[Ax-b], δx). \end{align*}
Hence, the derivative is 
.
That is because, $∇f = (∂_{e_1}f, ∂_{e_2}f, …)^T$. So replacing δx with 
gives: $$∂_{e_i} = {2A^\top[Ax-b]}_i.$$
Higher derivatives can be calculated in the same way: \begin{align*} \frac{d}{dε}(2A^\top[A(x+δxε-b])\big|_{ε=0} &= (2A^\top Aδx)\big|_{ε=0} \\ &=2A^\top Aδx \end{align*} $⇒∇^2f(x) = 2A^\top A.$
I'm doing some math warm-up questions for a fluids class I'm about to start and I'm stuck on a vector calculus question.
Matrix A is a 2x2 matrix with row 1 = [a b] and row 2 = [c d]. (Sorry, I can't figure out how to make a decent looking matrix, but I think you get the idea).
The exact question goes as follows: Show that ∇ · (∇ ∧ A) = 0. Here ∇ is the gradient operator in 2D and is given by ∇ = (∂/∂x)i + (∂/∂y)j.
I think the wedge operator is being used like a cross operator (×). Now, I understand that for a vector field, this is basically saying that the divergence of the curl is zero and that makes sense to me. Given a vector-valued function, I would just cross ∇ and the function and then take the dot product with what comes out of the cross product and find zero. What's throwing me off here is that A is a matrix and not a vector-valued function.
Could someone please help me understand what (∇ ∧ A) means? I've thought of just multiplying the ∇ vector with A and then dotting ∇ with the result, which gets me ∇A = [(∂/∂x)a + (∂/∂y)c]i + [(∂/∂x)b + (∂/∂y)d]j, and then ∇ · (∇A) = (∂/∂x)[(∂/∂x)a + (∂/∂y)c] + (∂/∂y)[(∂/∂x)b + (∂/∂y)d]. I don't think that's what I'm supposed to do, though. Help is much appreciated!