When using Lyapunov functions to prove stability and the Lyapunov function is dependent on time, eg:
$$V(x,t) = x_1\sin(t)$$
and I want to show that the derivative is negative $\dot{V}(x,t) < 0$,
Am I correct in also differentiating the Lyapunov function with respect to time t?
$$\dot{V}(x,t) = x_1\dot{x_1}+x_1\cos(t)$$
On
As stated by AVK, the answer would indeed be yes. However, I would also like to add some additional notes regarding the use of time varying Lyapunov functions. Namely, for time varying Lyapunov functions it is not enough to have $\dot{V}<0$ to show asymptotic stability. For example consider the following system
$$ \dot{x} = x $$
and Lyapunov function
$$ V(t,x) = x^2\,e^{-4t}, $$
such that
$$ \dot{V} = - 2\,x^2\,e^{-4t}, $$
which is negative definite in $x$. However, clearly the dynamics of $x$ is unstable.
Instead one can use Theorem 4.8 and 4.9 from Nonlinear Systems from Hassan K. Khalil, which roughly states that the dynamics $\dot{x} = f(t,x)$, with equilibrium point $x=0$, is asymptotically stable if
$$ W_1(x) \leq V(t,x) \leq W_2(x) \\ \dot{V} \leq - W_3(x) $$
with $W_1(x)$, $W_2(x)$ and $W_3(x)$ positive definite functions in $x$, i.e. for each $i\in\{1,2,3\}$ it hold that $W_i(0) = 0$ and $W_i(x) > 0\ \forall\,x\neq0$.
According to the chain rule $$ \dot V(x,t)= \frac{dV(x_1(t),\ldots,x_n(t),t)}{dt}= \frac{\partial V}{\partial x_1}\dot x(t)+\ldots+ \frac{\partial V}{\partial x_n}\dot x(t)+\frac{\partial V}{\partial t}\cdot 1 $$ So in your case the answer should be $$ \dot V(x,t)=\dot x_1\sin t+x_1\cos t $$