Unclear about matrix calculus in least squares regression The loss function of a Least Squares Regression is defined as (for example, in this question) :
$L(w) = (y - Xw)^T (y - Xw) = (y^T - w^TX^T)(y - Xw)$
Taking the derivatives of the loss w.r.t. the parameter vector $w$:
\begin{align}
\frac{d L(w)}{d w} & = \frac{d}{dw} (y^T - w^TX^T)(y - Xw) \\
 & = \frac{d}{dw} (y^Ty - y^TXw - w^TX^Ty + w^TX^TXw) \\
& = \frac{d}{dw} (y^Ty - y^TXw - (y^TXw)^T + w^TX^TXw)
\end{align}
as the second and third terms are scalars resulting in the same quantity, this implies,
\begin{align}
& = \frac{d}{dw} (y^Ty - 2y^TXw + w^TX^TXw)
\end{align}
My question is:
for the second term, shouldn't the derivative wrt $w$ be $-2y^TX$ ?
and because $\frac{d}{dx}(x^TAx) = x^T(A^T + A)$,
(see this question for explanation)
shouldn't the derivative for the third term (which is also a scalar), be the following due to chain rule? 
\begin{align}
\frac{d}{dw} (w^TX^TXw) + \frac{d}{dw} w(X^TXw)^T = w^T(X^TX + X^TX) = 2 w^TX^TX
\end{align}
From the above expressions, shouldn't the result of the derivative of the loss function be: $-2y^TX + 2 w^TX^TX$ ?
What I see in textbooks (including, for example, page 25 of this stanford.edu notes and page 10 of this harvard.edu notes ) is a different expression: $-2X^Ty + 2 X^TXw$. 
What am I missing here?
 A: Let $z=(Xw-y)$, then the loss function can be expressed in terms of the Frobenius norm or better yet, the Frobenius product as 
$$L=\|z\|^2_F = z:z$$
The differential of this function is simply
$$\eqalign{
 dL &= 2\,z:dz \cr
  &= 2\,z:X\,dw \cr
  &= 2\,X^Tz:dw \cr
}$$
Since $dL=\frac{\partial L}{\partial w}:dw,\,$ the gradient is
$$\eqalign{
 \frac{\partial L}{\partial w} &= 2\,X^Tz \cr
  &= 2\,X^T(Xw-y) \cr
}$$
The advantage of this derivation is that it holds true even if the vectors $\{w,y,z\}$ are replaced by rectangular matrices.
A: *

*Let $A=Xw-y$ and find the derivative map of the squared norm $L=\|A\|^{2}$:
$D_A\|A\|^{2}(H)=\left.\frac{d}{dt}\right|_{0}\|A+tH\|^{2}=\left.\frac{d}{dt}\right|_{0}\langle A+tH,A+tH\rangle=2\langle A,H\rangle$

*Use the chain rule $D_{w}(L\circ A)=D_{A}L\circ D_{w}A$ as follows,
$D_w\|A(w)\|^{2}(h)=\left.\frac{d}{dt}\right|_{0}\|A(w+th)\|^{2}=2\langle A,D_wA(h)\rangle$

*Arrive at the result: $2\langle Xw-y,Xh\rangle$

*Rewrite $2\langle Xw-y,Xh\rangle = 2\langle X^T(Xw-y),h\rangle$ and define the gradient vector $2X^T(Xw-y)$
