Chain Rule: Derivative of Squared Mahalanobis Distance I need to calculate the derivative of
\begin{equation}
d(\Pi',\Sigma^{-1})=\sum_{t=1}^{T}(Y_{t}-\Pi'X_{t})'\Sigma^{-1}(Y_{t}-\Pi'X_{t})
\end{equation}
Where $y_{t}=\left[\begin{array}{c} y_{1t} \\ \vdots \\ y_{kt}\end{array}\right]$
$X_{t}=\left[\begin{array}{c} 1 \\ x_{1t} \\ ... \\ x_{pt} \\ 
\end{array}\right]$
with $t=1,...,T$
and 
\begin{equation}
\Pi'=\left[\begin{array}{ccc} \mu & ... & \Phi_{1p} \\
\mu & ... & \Phi_{2k} \\ 
\vdots & \ddots & \vdots \\
 \mu & ... & \Phi_{kp}
\end{array}\right]
\end{equation}
The derivative would like 
\begin{equation}
\frac{\partial d(\Pi',\Sigma^{-1})}{\partial \Pi'}=2\sum_{t=1}^{T}\Sigma^{-1}(Y_{t}-\Pi'X_{t})\frac{\partial \Pi'X_{t}}{\partial \Pi'}\tag{*}
\end{equation}
What is the derivative of $\frac{\partial \Pi'X_{t}}{\partial \Pi'}$? 
I know that 
\begin{equation}
\frac{\partial \Pi'X_{t}}{\partial \Pi'}=\frac{\partial}{\partial \Pi'}\left[\begin{array}{c} \mu+\Phi_{11}X_{1t}+...+\Phi_{1p}X_{pt} \\
\vdots \\
\mu+\Phi_{1k}X_{1t}+...+\Phi_{kp}X_{pt}
\end{array}
\right]=\left[\begin{array}{c} 1 \\ X_{1t} \\ \vdots \\ X_{pt}\end{array}\right]
\end{equation}
With this result the dimensiones of the vectors in (*) do not match. I could transpose. But, I think the transposed must come naturally or not?
I have used the following proposition in the first factor of the previous derivative
Proposition
Let $\bf{x}$ a $n\times 1$ vector and $\bf{A}$ a $n\times n$ matrix such that $\bf{A'}=\bf{A}$. We define the function $q(\bf{x}): \mathbb{R}^{n}\rightarrow \mathbb{R}$ as
\begin{equation}
q(\bf{x})  =\bf{x'}\bf{A}\bf{x}
\end{equation}
Then 
\begin{equation}
\frac{\partial q(\bf{x})}{\partial x_{p}}=2\bf{A}\bf{x}
\end{equation}
First
\begin{equation}
\frac{\partial x_{k}}{\partial x_{p}}=\delta_{p,k}=
\begin{cases}
1 & \text{if}\quad k=p \\ 
0 & \text{if}\quad k\neq p 
\end{cases}
\end{equation}
Where $k,p=1,2,...,n$.
Then
\begin{align}
q(\bf{x}) & =\bf{x'}\bf{A}\bf{x}\\ 
& = \left[\begin{array}{ccc} x_{1} & ... & x_{n}  \end{array}\right]\left[\begin{array}{ccc} a_{11} &  & a_{1n} \\ 
& \ddots & \\ 
a_{n1} &  & a_{nn}
\end{array}\right]\left[\begin{array}{c} x_{1} \\ \vdots \\ x_{n}\end{array}\right] \\ 
& = \sum_{j=1}^{n}\sum_{i=1}^{n}a_{ij}x_{i}x_{j} \\ 
\end{align}
The gradient will be 
\begin{align}
\frac{\partial q(\bf{x})}{\partial x_{p}}& =\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}\frac{\partial }{\partial x_{p}}\left[x_{i}x_{j}\right] \\ 
& = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}\left[\frac{\partial x_{i}}{\partial x_{p}}x_{j}+x_{i}\frac{\partial x_{j}}{\partial x_{p}}\right] \\ 
& = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}\left[\delta_{i,p}x_{j}+\delta_{j,p}x_{i}\right] \\ 
& = \sum_{j=1}^{n}a_{ij}x_{j}+\sum_{i=j=1}^{n}a_{ij}x_{j} \\
& = 2\sum_{j=1}^{n}a_{ij}x_{j} \\ 
& = 2\left[\begin{array}{c} \sum_{j=1}^{n}a_{1j}x_{j} \\
\sum_{j=1}^{n}a_{2j}x_{j} \\
\vdots \\
\sum_{j=1}^{n}a_{nj}x_{j}
\end{array}\right] \\ 
& = 2\left[\begin{array}{ccc} a_{11} &  & a_{1n} \\ 
& \ddots & \\ 
a_{n1} &  & a_{nn}
\end{array}\right]\left[\begin{array}{c} x_{1} \\ \vdots \\ x_{n}\end{array}\right] \\
& = 2\bf{A}\bf{x}
\end{align}
 A: To reduce the visual clutter, let's use the Einstein summation convention and the variables
$$\eqalign{
P = \Pi',\quad S = \Sigma^{-1},\quad w_k=Px_k-y_k,\quad \lambda = d(\Pi',\Sigma^{-1})
}$$
and a colon to denote the trace/Frobenius product, i.e.
$$A:B={\rm Tr}(A^TB)$$
Write the distance. Then find its differential and gradient.
$$\eqalign{
\lambda &= w_k^TSw_k = S:w_kw_k^T \cr
d\lambda
 &= S:(w_k\,dw_k^T+dw_k\,w_k^T) \cr
 &= 2S:dw_k\,w_k^T \cr
 &= 2Sw_k:dw_k \cr
 &= 2Sw_k:dP\,x_k \cr
 &= 2Sw_kx_k^T:dP\cr
\frac{\partial\lambda}{\partial P}
 &= 2Sw_kx_k^T \cr
 &= 2\Sigma^{-1} \sum_{k=1}^T\big(\Pi'x_k-y_k\big)x_k^T \cr
}$$
Hopefully that last expression will make you realize that $\Sigma$ and $T$ are poorly chosen names for problems involving summations and transposes. Similarly, $\Pi'$ is a poor name if a product is involved.
A: Let $x$,$y$ be column vectors and $A$ $B$ square matrices. 
Then 
$$\begin{align}
\frac{d( [x + A y]^´ B 
[x + A y])}{dA}&= \frac{d (x^´ B^´ A y ) }{dA}+ \frac{d (y^´ A^{´}B A y ) }{dA}
+\frac{d (x^{'} B A y ) }{dA}\\
&=  Bx y' + B' A y y' + B A y y'  +B' x y' \\
&= (B+B') (x y' + y y')
\end{align}
$$
by properties $(70)$ and $(82)$ in the matrix cookbook
I think you can go on from here.
