We are learning matrix calculus in class. This is really new for me and I am trying to get extra practice by going through old assignments. One of the questions asks to find the derivative of the function below with respect to $\beta$ and if possible to find the value of the elements of $\beta$ that maximize the function. After taking logs to simplify the original function, I took the partial with respect to $\beta$. I thought my derivative was correct, but when I set the partial equal to zero, I am unable to solve for $\beta$. So either my derivative is wrong or I don't know enough to solve for $\beta$, or it's a trick question and there is no analytical expression for $\beta$. A more advanced student suggested that it can probably only be solved by numerical methods but I was wondering if anyone here can tell me if my derivative is correct and whether or not it's possible to solve for $\beta$?
$ x_{1} = \begin{pmatrix} a\\ b\\ \end{pmatrix}\quad x_{2} = \begin{pmatrix} c\\ d\\ \end{pmatrix}\quad \beta = \begin{pmatrix} \beta_{1}\\ \beta_{2}\\ \end{pmatrix}\quad Y = \begin{pmatrix} y_{1}\\ y_{2}\\ \end{pmatrix}\quad $
f(x_{1},x_{2},y_{1},y_{2},\beta) = \frac{1}{x_{1}'\beta}\exp\left\{\frac{-y_{1}}{x_{1}'\beta}\right\}\frac{1}{x_{2}'\beta}\exp\left\{\frac{-y_{2}}{x_{2}'\beta}\right\}
My answer:
\frac{\partial \ln f}{\partial\beta}=-\frac{1}{x_{1}'\beta}x_{1}-\frac{1}{x_{2}'\beta}x_{2}+\frac{x_{1}y_{1}}{x_{1}'\beta.x_{1}'\beta}+\frac{x_{2}y_{2}}{x_{2}'\beta.x_{2}'\beta}=0
Is this correct and is it possible to solve for $\beta$? If so, how? Thanks!