How to solve:
$${\frac{\partial}{\partial{\vec{\mu}^T}}\left\{\frac{1}{a}\sum^n_{i=1}z_i(\vec{y_i}-\vec{\mu})^TB^{-1}\right\}}$$
where $a$ and $z_i$ are elements, $\vec{y_i}$ and $\vec{\mu}$ are both $2\times1$ vectors, $B$ is a $2\times2$ matrix.
And is there any software can do such general differentiations?
Well, you can use the matrix calculus notation if you'd like to. Anyways, you have a derivative of a sum so you can turn it into a sum of derivatives (as I understood $a, \vec{z}, \vec{y_i} $ and $ B$ (or $ B^{-1}$, that doesn't matter) do not depend on $\vec{\mu}$): $$ \begin{eqnarray} {\frac{\partial}{\partial{\vec{\mu}^T}}\left\{\frac{1}{a}\sum^n_{i=1}z_i(\vec{y_i}-\vec{\mu})^TB^{-1}\right\}}&=&{\frac{\partial}{\partial{\vec{\mu}^T}}\left\{\frac{1}{a}\sum^n_{i=1}z_i(\vec{y_i}^TB^{-1})-\frac{1}{a}\sum^n_{i=1}z_i(\vec{\mu}^TB^{-1})\right\}}\\ &=&\frac{\partial}{\partial{\vec{\mu}^T}}\left\{-\frac{1}{a}\sum^n_{i=1}z_i(\vec{\mu}^TB^{-1})\right\}\\ &=&-\frac{1}{a}\sum^n_{i=1}z_i\frac{\partial}{\partial{\vec{\mu}^T}}\vec{\mu}^TB^{-1}\\ &=&-\frac{1}{a}\sum^n_{i=1}z_i (B^{-1})^T \end{eqnarray} $$