The path integral propagator for a free non-relativistic particle in one dimension is,
$$K(x_1,T;x_0,0)=\int_{x(0)=x_0}^{x_1(T)=x_1}\mathcal{D}[x]\exp\left[\frac{i}{\hbar}\int_0^T\mathrm{d}t\frac{1}{2}m\dot{x}^2\right]=\sqrt{\frac{m}{2\pi i\hbar T}}\exp\left[\frac{i}{\hbar}S_{\rm classical}\right].$$
However, now I am dealing with a case where the mass is time dependent, $$K(x_1,T;x_0,0)=\int_{x(0)=x_0}^{x_1(T)=x_1}\mathcal{D}[x]\exp\left[\frac{i}{\hbar}\int_0^T\mathrm{d}t\frac{1}{2}M(t)\dot{x}^2\right].$$
So I was wondering what the propagator becomes in this case? Does it retain a form like $$\sqrt{\frac{M(t)}{2\pi i\hbar T}}\exp\left[\frac{i}{\hbar}S_{\rm classical}\right],$$
where the $S_{\rm classical}$ changes to
$$S_{\rm classical}=\frac{im}{2\hbar}\frac{\left(x_1-x_0\right)^2}{T}\rightarrow S_{\rm classical}=\int_0^T{\rm d}t\frac{1}{2}M(t)\left(\frac{-M(t)\pm\sqrt{M^2(t)-c\dot{M}(t)}}{\dot{M}(t)}\right)^2,$$
because the classical velocity satisfies the following equation,
$$\frac{{\rm d}}{{\rm d}t}\left(\dot{M}(t)\dot{x}^2+2M(t)\dot{x}\right)=0.$$
Can someone answer? It would be better if one shows the derivation.