Differentiation of a function including vector

Question

$\beta$ and $x$ are $r\times 1$ vectors. $\Sigma$ is a $r\times r$ symmetric matrix. $f(\cdot)$ is a function of $\frac{\beta^Tx}{(\beta^T\Sigma\beta)^{1/2}}$. How do we get the following quantity? \begin{eqnarray*} \frac{\partial^2f}{\partial\beta\partial\beta^T}= \frac{\partial^2}{\partial\beta\partial\beta^T}f\left(\frac{\beta^Tx}{(\beta^T\Sigma\beta)^{1/2}}\right).\end{eqnarray*} It would be appreciated if you let me know how to solve.

Answer 1

Use Chain rule, call $t = \frac{\beta^T x}{(\beta^T \Sigma \beta)^{1/2}}$

The first derivative is

\begin{equation} \frac{\partial f(t)}{\partial \beta} = \frac{\partial f(t)}{\partial t} \frac{\partial t}{\partial \beta} \end{equation} The first term depends on the function as for the second term: \begin{equation} \frac{\partial t}{\partial \beta} = \frac{\partial}{\partial \beta} \Big( \frac{\beta^T x}{(\beta^T \Sigma \beta)^{1/2}} \Big) = \frac{(\beta^T \Sigma \beta)^{1/2}x - \frac{\beta^Tx}{\sqrt{\beta^T\Sigma\beta}}\Sigma\beta}{(\beta^T \Sigma \beta)} \end{equation} which could be further simplified. I leave the second derivative to you.