Derivative of a time-varying Lyapunov function

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 $$ \dot V(x,t)=\dot x_1\sin t+x_1\cos t $$

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 $W_i(x) > 0\ \forall\,x\neq0$.