Let $z(t)=[z_1(t),z_2(t)]^T$
Also, $V(z)=az_1^2+2bz_1z_2+cz_2^2$
I am trying to find $V'=\frac{d}{dz} V$. I know this means I have to take partials with respect to $z_1$ and $z_2$, but the middle term always throws me off.
I know for $V(z)=z_1^2+z_2^2, V'=2z_1z_1'+2z_2z_2'$, but I'm having trouble remembering what the partial of $2bz_1z_2$ is specifically. Is it $2b(z_1+z_2)$ or $2b(z_1z_2'+z_2z_1')$?
I might have some notation wrong, but the second example I know is correct.
The derivative you are seeking is a Jacobian of $V$:
$$V'={d\over dz}V=\begin{pmatrix} {\partial V\over \partial z_1} & {\partial V\over \partial z_2} \end{pmatrix}$$
where the partials are:
$$\begin{align}{\partial V\over \partial z_1} &= 2az_1+2bz_2 \\ {\partial V\over \partial z_2} &= 2bz_1+2cz_2\end{align}$$
I have a clue you are looking for ${d\over dt}V$, which requires application of a chain rule:
$$V''={d\over dt}V = {dV\over dz}\cdot{dz\over dt} = \begin{pmatrix} {\partial V\over \partial z_1} & {\partial V\over \partial z_2} \end{pmatrix}\cdot\begin{pmatrix} {\partial z_1\over \partial t} \\ {\partial z_2\over \partial t}\end{pmatrix} = {\partial V\over \partial z_1}\cdot{\partial z_1\over \partial t} + {\partial V\over \partial z_2}\cdot{\partial z_2\over \partial t}$$
Be aware that ${dV\over dz}$ is computed at $z(t)$ while ${dz\over dt}$ is computed at $t$.