The answer is correct and could be simplified as$$\frac{d}{dt}\frac{1}{|\vec{x}(t)|} =$$
$$ -x_1(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_1}{dt} -x_2(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_2}{dt} -x_3(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_3}{dt}=$$
$$-(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}} (x_1\frac{dx_1}{dt}+x_2\frac{dx_2}{dt}+x_3\frac{dx_3}{dt})=$$
$$- \frac{1}{|\vec{x}|^3} \left(\vec{x} \cdot \frac{d \vec{x}}{dt} \right).$$
"Sum over paths" is 150% correct. It, in fact, means product over integrals over each point of a path.
A picture should be worth a thousand words. Discretize the integral by two screens, with 3 and 4 slits, respectively. Ultimately, you'll consider the limit of an infinity of screens, each with an infinity of slits.
Think of your assignment to go from A to B in straight segments through one of the 3 slits at t1, and then the 4 at t2. You have then 3×4=12 options, 12 paths, 12 histories to achieve that. Your first (topmost) sum will then have 12 terms of exponentials of $iS_1, ..., iS_{12}$, the sum of 12 complex amplitudes, each corresponding to a different path. But each path/amplitude is a product of three segments, characterized by their endpoint: that 's why we are skipping the last factor of the segment abutting at B.
Now, each path/history is characterized by the location of its slit on each screen; e.g., calling each slit location 1,2,3, and 1,2,3,4, respectively,
$$
S_i=\int dt ~ L_i(x(t)) \leftarrow L(x_1(i)) + L(x_2(i)) \leadsto \\
S_1 = L(x_1=1) +L(x_2=1) , \\
S_2 = L(x_1=2) +L(x_2=1) , \\
S_3 = L(x_1=3) +L(x_2=1) , \\
S_4 = L(x_1=1) +L(x_2=2) ,... \\
S_{12} = L(x_1=3) +L(x_2=4) .
$$
The sum of the 12 exponentials then amounts to
$$
\sum_{x_1} \sum_{x_2} e^{i L(x_1) +iL(x_2)}= \sum_{x_1} \sum_{x_2} e^{i L(x_1)} e^{iL(x_2)},
$$
a sum over 12 paths.
It should then be evident how to generalize to the continuum path integral.
First, still keeping just two screens, but perforating them with an infinity of slits to the point of evanescence, you get
$$
\sum_{x_1} \to \int dx_1 , \sum_{x_1} \to \int dx_1 .
$$
You then take the number of screens to N and then infinity, so
$$
L(x_1) +L(x_2) \to L(x_1) +L(x_2)+...+ L(x_N) \to \int dt ~L(x(t)),
$$
which, times i, is exponentiated to provide the integrand of the functional integral, now over an infinity of screens, $\int dx_1 \int dx_2 ...\int dx_N \to \iint \cal Dx $.
In sum,
$$
\iint \cal{D} x ~~ e^{i\int dt ~L(x(t))}.
$$
You might review the free particle to fix your notation.
- Τhis is, of course, all in black and white on pp 68-69 of Dirac's epochal paper where he introduces the path integral Feynman developed.
Best Answer
The answer is \begin{equation} \frac{\partial f}{\partial a_2}=\frac{1}{\sqrt{\vert T \vert} (2\pi)^{3/2}} \int_{-\infty}^{x_3=a_3} dx_3 \int_{-\infty}^{x_1=a_1} dx_1 \begin{bmatrix}{x_1\\a_2\\x_3}\end{bmatrix} exp\bigg(-0.5\begin{bmatrix}{x_1 a_2 x_3}\end{bmatrix} T^{-1} \begin{bmatrix}{x_1\\a_2\\x_3}\end{bmatrix} \bigg). \end{equation}
Consider the antiderivative $F(x)$ of the function \begin{equation} F(x) = \int f(x) dx \end{equation} or \begin{equation} \frac{d F(x)}{dx} = f(x), \end{equation} so definite integral is the difference of antiderivatives \begin{equation} \int\limits_a^b f(x)dx = F(b) - F(a), \end{equation} i.e. the derivative is \begin{equation} \frac{\partial}{\partial b} \int\limits_a^b f(x)dx = F^{\prime}(b) = f(b). \end{equation}