Help with matrix calculus? 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!
 A: This function doesn't have a global maximum, at least not if $x_1$ and $x_2$ are linearly independent: You can choose $x_1'\beta$ such that the first exponential goes to $\infty$, and you can independently choose the sign of $x_2'\beta$ such that the product is positive and thus goes to $\infty$.
However, you can still look for the stationary points and see whether they are local maxima. Taking the logarithm before taking the derivative is a good approach, and the resulting equation can be solved with nothing but linear algebra.
If $x_1$ and $x_2$ are linearly independent, their coefficients in this equation have to vanish individually. That gives you one linear equation for $x_1'\beta$ and one for $x_2'\beta$. You can solve these, and then knowing the scalar products of $\beta$ with two linearly independent vectors, you can reconstruct $\beta$.
If $x_1$ and $x_2$ aren't linearly independent, you can substitute $x_2=\lambda x_1$ into the equation, and setting the coefficient of $x_1$ in the result to $0$ gives you a linear equation for $x_1'\beta$. In this case the stationarity condition only restricts $\beta$ in one direction, since the function is independent of the component of $\beta$ orthogonal to $x_1$ and $x_2$.
