Feynman's Quantum Mechanics and Path Integral has a vivid physical interpretation of the path integral formalism. But I was stumbled on some mathematical details while following his derivation of the Schrodinger equation of "kernel" from the "kernel integral equation". A kernel $K(b,a)=K(x_b,t_b;x_a,t_a)$ is defined as a path integral between two endpoints $(x_b,t_b)$ and $(x_a,t_a)$ with the time instances $t_b<t_a$. It follows the relation of
$$K(b,a)=\int_{-\infty}^{\infty}K(b,c)K(c,a)dx_c,\tag{1}$$
with the Schrodinger equation for the kernel being
$$\frac{\partial}{\partial t_b}K(b,a)=\frac{-i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K(b,a)+V(x_b,t_b)K(b,a)\right]=-\frac{i}{\hbar}\hat{H}_bK(b,a).\tag{2}$$
$V(x_b,t_b)$ in eqn.(2) is a potential function. Because $K(b,a)=0$ for $t_b<t_a$, and
$$\lim_{t_b\rightarrow t_a+0^+}K(b,a)=\delta(x_b-x_a),\tag{3}$$
Feynman then makes eqn. (2) complete by writing
$$\frac{\partial}{\partial t_b}K(b,a)=\frac{-i}{\hbar}\hat{H}_bK(b,a)+\delta(x_b-x_a)\delta(t_b-t_a).\tag{4}$$
Now, if we treat the potential $V(b)=V(x_b,t_b)$ as a perturbation to a system represented by $K(b,a)$, and write
$$K(b,a)=\int_a^bexp\left\{\frac{i}{\hbar}\int_{t_a}^{t_b}(\frac{m}{2}\dot{x}^2-V(x,t))dt\right\}\mathfrak{D}x(t)\tag{5}$$
and
$$K_0(b,a)=\int_a^bexp\left\{\frac{i}{\hbar}\int_{t_a}^{t_b}\frac{m}{2}\dot{x}^2dt\right\}\mathfrak{D}x(t)\tag{6},$$
we can expand $exp\{-\frac{i}{\hbar}\int_{t_a}^{t_b}V(x,t)dt\}$ in eqn.(5) to give
$$K(b,a)=K_0(b,a)-\frac{i}{\hbar}\int K_0(b,c)V(c)[K_0(c,a)-\frac{i}{\hbar}\int K_0(c,d)V(d)K_0(d,a)d\tau_a+\dots]d\tau_c\tag{7}$$
which gives the integral equation:
$$K(b,a)=K_0(b,a)-\frac{i}{\hbar}\int K_0(b,c)V(c)K(c,a)d\tau_c\tag{8},$$
where $d\tau=dxdt$. In the book, Feynman argued that the kernel $K(b,a)$ satisfies the following equation:
$$\frac{\partial}{\partial t_b}K(b,a)+\frac{i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K(b,a)+V(x_b,t_b)K(b,a)\right]=\delta(t_b-t_a)\delta(x_b-x_a)\tag{9},$$
which is just eqn.(4). Now, in the problem 6-3 of the book it says that the eqn.(9) can be derived by using "free particle Schrodinger equation",
$$\frac{\partial}{\partial t_b}K_0(b,a)+\frac{i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K_0(b,a)\right]=\delta(t_b-t_a)\delta(x_b-x_a),\tag{10}$$
and the integral equation (8). To do this, I simply take a derivative w.r.t. $t_b$ on both size of eqn.(8), and notice that
$$\int K_0(b,c)V(c)K(c,a)d\tau_c=\int_{t_a}^{t_b}\int_{-\infty}^{\infty}K_0(b,c)V(c)K(c,a)dx_cdt_c\tag{11}$$
should also be treated as a function of $t_b$. So I applied the Leibniz integral rule to the second term on the RHS of eqn.(8) to find
$$\frac{\partial}{\partial t_b}K(b,a)=\frac{\partial}{\partial t_b}K_0(b,a)-\frac{i}{\hbar}\int_{-\infty}^{\infty}K_0(x_b,t_b;x_c,t_b)V(c)K(c,a)dx_c-\frac{i}{\hbar}\int \frac{\partial}{\partial t_b}K_0(b,c)V(c)K(c,a)d\tau_c.\tag{12}$$
I use eqn.(3) in the second term above and replace $\partial K_0/\partial t_b$ with the terms in eqn. (10) I further find that
$$\frac{\partial}{\partial t_b}K(b,a)=\frac{\partial}{\partial t_b}K_0(b,a)-\frac{i}{\hbar}V(b)K(b,a)-\frac{i}{\hbar}\frac{i}{\hbar}\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}\int K_0(b,c)V(c)K(c,a)d\tau_c-\frac{i}{\hbar}V(b)K(b,a).\tag{13}$$
Because $K_0(b,a)-K(b,a)=\frac{i}{\hbar}\int K_0(b,c)V(c)K(b,a)d\tau_c$, I finally arrive at
$$\frac{\partial}{\partial t_b}K(b,a)+\frac{i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K(b,a)+2V(x_b,t_b)K(b,a)\right]=\delta(t_b-t_a)\delta(x_b-x_a).\tag{14}$$
Now, eqn.(14) is very much like eqn.(9), but the factor 2 in front of $V(x_b,t_b)$ really confused me. Can anybody tell me what's wrong with my derivation leading to eqn.(14)? Any help will deeply appreciated.
