I have the equation as
$$F(w,x)=\sum_{i=1}^{N}\int_{x \in \Omega} \left ( Y(x)-w^TA(x)\right)^2u_i(x)dx$$
In which, $w$ is column vector that independent on $x$, denotes $w=[w_1,w_2...,w_M]^T$
$A$ is column vector value that denotes $A(x)=[a_1(x),a_2(x),...,a_M(x)]^T$
$(.)^T$ is transpose operator. $Y(x)$ and $u_i(x)$,$A(x)$ are independent on $w$.
I want to perform derivative of $F(x,w)$ with respect to $w$. Could you see my answer and give me some suggestion that Is it correct? $$\frac {\partial F(x,w)}{\partial w}=?$$
That is what I done
we see that $w^TA(x)=\sum_{i=1}^{M} w_ia_i$, then
$$\frac {\partial (w^TA)}{\partial w_i}=ai$$
For $i=1$ to $M$, hence $$\frac {\partial (w^TA)}{\partial w}=A$$
Hence, $$\frac {\partial F(w,x)}{\partial w}=-2\sum_{i=1}^{N}\int_{x \in \Omega} \left ( Y(x)-w^TA(x)\right)u_i(x)\frac {\partial (w^TA)}{\partial w}dx$$ $$\frac {\partial F(w,x)}{\partial w}=-2\sum_{i=1}^{N}\int_{x \in \Omega} \left ( Y(x)-w^TA(x)\right)u_i(x)A(x)dx$$
$$\frac {\partial F(w,x)}{\partial w}=-2\sum_{i=1}^{N}\int_{x \in \Omega} \left ( Y(x)-w^TA(x)\right)A(x)u_i(x)dx$$
Because $w^TA(x)=A(x)w$ then
$$\frac {\partial F(w,x)}{\partial w}=-2\sum_{i=1}^{N}\int_{x \in \Omega} \left ( Y(x)A(x)-A(x)A^T(x)w\right)u_i(x)dx$$ I'm wondering about the last two steps.
You have to use this way to get the derivative: \begin{eqnarray} D_wF(h)(w,x)&=&\frac{d}{dt}F(w+th,x)\bigg|_{t=0}\\ &=&\sum_{i=1}^{N}\int_{x \in \Omega} \frac{d}{dt}\left[(Y(x)-(w+th)^TA(x))^2\right]u_i(x)\bigg|_{t=0}dx\\ &=&-2\sum_{i=1}^{N}\int_{x \in \Omega} (Y(x)-w^TA(x))h^TA(x)u_i(x)dx. \end{eqnarray}