Second order derivative of log of vector I have a vector of size $n$ x $1$ named $\alpha$. Let $f(\alpha) = u\cdot\mathbf 1^{\!\top}ln(\alpha)$ where $u$ is scalar.
What is the $f'(\alpha)$ and $f''(\alpha)$ and equivalent Matlab code?
According to me the first derivative is
$$f'(\alpha) = u/\alpha$$
and equivalent MATLAB code is --
f_a_1 = u ./ a
and for the second derivative
$$f''(\alpha) = u\cdot(Diag(\alpha)*Diag(\alpha))^{-1}$$
Equivalent MATLAB code is
f_a_2 = u*inv(diag(a)*diag(a))
Is my inference correct?
 A: Let $\alpha =(\alpha_1,\alpha_2,\dots,\alpha_n)$
You probably (see the note below to understand the doubts) defined function $f:\mathbb{R}^n\to \mathbb{R}$
$$f(\alpha)=f(\alpha_1,\alpha_2,\dots,\alpha_n)=u\sum_{i=1}^n \ln\alpha_i$$
therefore 
If you need a vector gradient of $f$ (a vector of partial derivatives), then you denote it as $$\nabla f=\left(f_{\alpha_1},\dots,f_{\alpha_n}\right)=\left(\frac{u}{\alpha_1},\dots,\frac{u}{\alpha_n}\right)$$
and compute it in matlab as 

u./a

if you looking for a total derivative of $f$
it is defined as $\nabla f\cdot \alpha$ and in your case is equal to $u n$
and another differentiation will be $0$. 

You can do

-u./(a.^2)

in matlab (without converting it to diagonal matrices etc), but this is not a second derivative of your function.
I would say you have to really clarify your question.

Note, the $\ln$ of a matrix is defined for $n\times n$ matrices, so the notatoins of $\ln$ of vector are incorrect and misleading.
The truth is that 
$$\exp{
\begin{bmatrix}
a_{11}& \cdots &a_{1n}\\
\vdots & \ddots & \vdots\\
a_{n1} & \cdots &a_{nn}
\end{bmatrix}
}\ne
\begin{bmatrix}
e^{a_{11}}& \cdots &e^{a_{1n}}\\
\vdots & \ddots & \vdots\\
e^{a_{n1}} & \cdots &e^{a_{nn}}
\end{bmatrix}
$$
neigher
$$\ln{
\begin{bmatrix}
a_{11}& \cdots &a_{1n}\\
\vdots & \ddots & \vdots\\
a_{n1} & \cdots &a_{nn}
\end{bmatrix}
}\ne
\begin{bmatrix}
\ln{a_{11}}& \cdots &\ln{a_{1n}}\\
\vdots & \ddots & \vdots\\
\ln{a_{n1}} & \cdots &\ln{a_{nn}}
\end{bmatrix}
$$
They are acutally defined trough power series of $\ln$ and exponential.
See link and link
However in matlab the regular 

exp

and 

log

do an elementwise evaluation of matrix and vector entries, e,g. 

exp([a,b,c])

will return the value of 

[exp(a),exp(b),exp(c)].

In matlab, the true matrix $\ln$ and exponential implemented via 

logm

and

expm

