Differentiation of multiple correlation coefficient Hi is there anyone that can help me with this calculation? 
So, I don’t know the exact way to differentiate the fraction of vector function. I tried to use the formal way but the components of the numerator had different size of dimension. Btw I actually solved the question by Cauchy-Schwarts inequality but am just curious with the technique the solution suggested. Thanks.
 A: From the derivation,
$$\rho^2_{yw} = \frac{\left(\boldsymbol\alpha^\mathsf T 
\boldsymbol \sigma_{yx}  \right)^2}{   \sigma_{yy} \left(\boldsymbol\alpha^\mathsf T 
\boldsymbol \Sigma_{xx} \boldsymbol \alpha \right)}\tag 1.$$
The objective is to seek the maximum squared correlation between $y$ and any linear function of $\mathbf x,$ say $\boldsymbol\alpha^\mathsf T \mathbf x. $
The author then shows how to proceed: by finding the value of $\boldsymbol\alpha$ that maximises $\rho^2_{yw},$ i.e. one needs to find $\boldsymbol\alpha $ such that
$$\frac{\partial}{\partial  \boldsymbol\alpha} \rho^2_{yw}=\mathbf 0.$$
Equivalently, one needs to find $\boldsymbol\alpha $ such that
$$\frac{\partial}{\partial  \boldsymbol\alpha} \ln(\rho^2_{yw})= \mathbf 0.$$
From $(1) , $
\begin{align}\frac{\partial}{\partial  \boldsymbol\alpha} \ln(\rho^2_{yw}) &=  \frac{\partial}{\partial  \boldsymbol\alpha} \left[2\ln \left(\boldsymbol\alpha^\mathsf T 
\boldsymbol \sigma_{yx}  \right)-\ln\sigma_{yy}-\ln\left(\boldsymbol\alpha^\mathsf T 
\boldsymbol \Sigma_{xx} \boldsymbol \alpha \right)\right]\\ &\overset{(\star), (\star\star)}{=}\frac2{\boldsymbol\alpha^\mathsf T 
\boldsymbol \sigma_{yx}}\cdot \boldsymbol \sigma_{yx}- \frac1{\boldsymbol\alpha^\mathsf T 
\boldsymbol \Sigma_{xx} \boldsymbol \alpha }\cdot 2\boldsymbol \Sigma_{xx} \boldsymbol \alpha,\tag 2
  \end{align}
where
\begin{align}
(\star):\frac{\partial\left(\mathbf a^\mathsf T\mathbf x\right) }{\partial \mathbf x} &= \mathbf a, \\(\star\star):\frac{\partial\left(\mathbf x^\mathsf T\mathbf A\mathbf x\right) }{\partial \mathbf x} &= 2\mathbf A\mathbf x, 
\end{align}
$\mathbf A$ being symmetric.
Now, from $(2), $
\begin{align}
\frac{\partial}{\partial  \boldsymbol\alpha} \ln(\rho^2_{yw}) &= \mathbf 0\\ \implies \frac2{\boldsymbol\alpha^\mathsf T 
\boldsymbol \sigma_{yx}}\cdot \boldsymbol \sigma_{yx}- \frac1{\boldsymbol\alpha^\mathsf T 
\boldsymbol \Sigma_{xx} \boldsymbol \alpha }\cdot 2\boldsymbol \Sigma_{xx} \boldsymbol \alpha &= \mathbf 0\\ \implies \frac{\boldsymbol \Sigma^{-1}_{xx}\boldsymbol \sigma_{yx}}{\boldsymbol\alpha^\mathsf T 
\boldsymbol \sigma_{yx}}- \frac{   \boldsymbol\alpha}
   {\boldsymbol\alpha^\mathsf T 
\boldsymbol \Sigma_{xx} \boldsymbol \alpha }&= \mathbf 0 \\ \implies \boldsymbol \alpha &= \left( \frac{\boldsymbol\alpha^\mathsf T 
\boldsymbol \Sigma_{xx} \boldsymbol \alpha }{\boldsymbol\alpha^\mathsf T 
\boldsymbol \sigma_{yx}} \right)    \boldsymbol \Sigma^{-1}_{xx}\boldsymbol \sigma_{yx}.\end{align}
Now, value of $\boldsymbol\alpha$ has to be substituted in $(1) $ to get the $\max_{\alpha}\rho^2_{yw}.$
