Originally Posted by

**choovuck** no, the Hessian matrix has as its (i,j)-th element the derivative $\displaystyle \frac{\partial^2 f}{\partial x_i \partial x_j}$, so in our case

$\displaystyle

HessianMatrix=

\begin{bmatrix}

\frac{\partial^2{f(x, \alpha)}}{\partial{x}^2}

&\frac{\partial^2{f(x, \alpha)}}{\partial{x}\partial{\alpha}} \\

\frac{\partial^2{f(x, \alpha)}}{\partial{\alpha}\partial{x}}

&\frac{\partial^2{f(x, \alpha)}}{\partial{\alpha}\partial{\alpha}}

\end{bmatrix}

$

It is symmetric since $\displaystyle \frac{\partial^2{f(x, \alpha)}}{\partial{\alpha}\partial{x}}=\frac{\part ial^2{f(x, \alpha)}}{\partial{x}\partial{\alpha}}$

I'm not very comfortable with those matrix gradients (I can't even convince myself why your $\displaystyle \frac{\partial f}{\partial \alpha}$ is correct...), so I won't help you with the actual computations, sorry