How to show that $\partial S/\partial q=p$ without variation of $S$? I'm trying to get some understanding in treating action $S$ as a function of coordinates. Landau and Lifshitz consider $\delta S$, getting $\delta S=p\delta q$, thus concluding that
$$\frac{\partial S}{\partial q}=p.$$
I'm now trying to understand this result, and consider the definition of action. I suppose that action as a function of coordinates $x$ will have $q=q(x,t)$ and $\dot q=\dot q(x,t)$, so $S(x)$ will look like:
$$S(x)=\int_{t_1}^{t_2}L(q(x,t),\dot q(x,t),t)dt.$$
Now I take the partial derivative with respect to $x$ to get $p$:
$$\frac{\partial S}{\partial x}=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\frac{\partial q}{\partial x}+\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial x}\right)dt.$$
And now I'm stuck. I see the momentum $\frac{\partial L}{\partial \dot q}=p$ and force $\frac{\partial L}{\partial q}=F$ inside the integral, but I can't seem to get, how to extract the momentum, so that all the other things cancelled.
Am I on the right track? What should be the next step?
 A: Comments to the question (v1): 


*

*The main point is that Ref. 1 is considering the (Dirichlet) on-shell action function
$$ \tag{1} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl}],  $$
not the (off-shell) action functional
$$\tag{2} I[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ L(q(t),v(t),t)\right|_{v(t)=\dot{q}(t)}. $$
See e.g. this Phys.SE post.

*The Lagrangian canonical/conjugate momentum is defined as
$$\tag{3}  p(t)~:=~\frac{\partial L(q(t),v(t),t)}{\partial v(t)},$$
as OP mentions.

*The statement that OP wants to prove is:

$$\tag{4}  \frac{\partial S}{\partial q_f}~=~p(t_f) , \qquad   \frac{\partial S}{\partial q_i}~=~-p(t_i).$$


*Equation (4) is e.g. proven in my Phys.SE answer here. 
References:


*

*L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$ 43.

A: Thanks to Qmechanic's answer, I've understood that $\partial S/\partial x=p(t_2)$, while I was under the illusion that it would somehow equal $p(t)$.
Now follows the finalization of my attempts.
First, consider the second part of the expression inside the integral for $\partial S/\partial x$ in the OP,
$$A=\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial x}dt.$$
Integrating by parts, namely using $u=\frac{\partial L}{\partial\dot q}$, $dv=\frac{\partial\dot q}{\partial x}dt$, we get, with $du=\frac d{dt}\frac{\partial L}{\partial\dot q}dt=\frac{\partial L}{\partial q}dt$ and $v=\frac\partial{\partial x}\int\dot q dt=\frac{\partial q}{\partial x}$,
$$A=\left.p\frac{\partial q}{\partial x}\right|_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{\partial q}{\partial x}\frac{\partial L}{\partial q}dt.$$
Substituting this into the integral in the OP, we get the integrals cancel, thus
$$\left.\frac{\partial S}{\partial x}=p\frac{\partial q}{\partial x}\right|_{t_1}^{t_2}=p(t_2)\frac{\partial q(x,t_2)}{\partial x}-p(t_1)\frac{\partial q(x,t_1)}{\partial x}.$$
But by definition of $x$, $q(x,t_2)\equiv x$, and because first point is fixed, we have $q(x,t_1)=q^{(1)}=\text{const}(x)$, thus the result is what was to be proved:
$$\frac{\partial S}{\partial x}=p(t_2).$$
