Suppose the quadratic form
$$V(x(t), t) = \frac{1}{2} x^\mathsf{T}(t) P(t) x(t)$$
where
$$x(t) \in \mathbb{R}^n,~P(t) \in \mathbb{R}^{n \times n},~\text{and}~P(t) = P^\mathsf{T}(t) > 0$$
i.$\,$e. $P(t)$ is a symmetric and positive definite matrix and $t$ denotes the time as an independed variable.
How does one determine $\frac{\partial V(x(t), t)}{\partial t}$?
We have
$$\frac{\partial}{\partial t} \langle x(t),P(t)x(t)\rangle=\frac{\partial}{\partial t}\sum_{i,j=1}^nP_{i,j}(t)x_i(t)x_j(t) = \sum_{i,j=1}^n\left(x_i(t)x_j(t)\frac{\partial}{\partial t}P_{i,j}(t)+x_j(t)P_{i,j}(t)\frac{\partial}{\partial t}x_i(t)+x_i(t)P_{i,j}(t)\frac{\partial}{\partial t}x_j(t)\right)$$
In particular if $\left(\frac{\partial}{\partial t}x(t)\right)_i=\frac{\partial}{\partial t}x_i(t)$ and $\left(\frac{\partial}{\partial t}P(t)\right)_{i,j}=\frac{\partial}{\partial t}P_{i,j}(t)$ for every $i,j=1,\ldots,n$, then
$$\frac{\partial}{\partial t} \langle x(t),P(t)x(t)\rangle=2\left\langle \left(\frac{\partial}{\partial t}x(t)\right),P(t)x(t)\right\rangle+\left\langle x(t),\left(\frac{\partial}{\partial t}P(t)\right)x(t)\right\rangle$$
where we used the symmetry $P(t)$.