Understanding Derivation of Euler Lagrange I am trying to understand the derivation of the Euler-Lagrange equation. I drew a graph below.

So, according to the graph,
$$ \int_{t_1}^{t_2} L(x+\delta{x},\dot{x}+\delta\dot{x}\,t)  dt - \int_{t_1}^{t_2} L(x,\dot{x},t) dt = \delta{S}$$
That is, we take the difference between the integral of the true path and the integral of the variated path, where S is the action defined by $\int_{t_1}^{t_2} L(x,\dot{x},t) dt$ and $\delta{S}$ is the variation in action (in the graph it is the distance between the blue line and the red line at any t value). Where do I proceed from there? I don't understand this line in the derivation:
$$ \int_{t_1}^{t_2} L(x,\dot{x},t) dt + \frac{d{L}}{d{x}}\delta{x} + \frac{d{L}}{d{\dot{x}}}\delta{\dot{x}}+O((\delta{x})^2)- L(x,\dot{x},t) dt = \delta{S}$$
(Source: http://wiki.math.toronto.edu/TorontoMathWiki/index.php/Euler-Lagrange_Equation)
Namely, what is the function O? Where are the partial derivatives coming from? The only thing I know to do is to combine the two integrals because the times ($t_1$ and $t_2$) and the integration variables (dt) are the same. How do I get all the mess to the left of $L(x,\dot{x},t)$ in the equation above?
 A: First O(...) is known as "big O" and represents the order of magnitude of the argument. 
Quick example:$F(x)= 1+x+x^2\rightarrow O(F(x))=O(x^2)$. That is to say the the limiting behaviour of the function F(x) is $x^2$.
To get the E-L equations you want to take the Taylor series of your Lagrangian L. 
Recap of 1D, the Taylor series is the expansion of a function in terms of the derivatives.
$$F(x)=\Sigma_{i=0}^\infty \frac{d^i}{dx^i}F(x)|_{x=x_0}(x-x_0)^i $$
This is meant to expand around some critical point $x_0$ but most of the time we take $x_0=0$. Further, let's look at a point $x=x_0+\Delta x$. The first couple of terms are as follows.
$$F(x)=F(x_0)+ \frac{d}{dx}F(x_0) \Delta x + \frac{d^2}{dx^2}F(x_0) \Delta x^2$$
In terms of big O notation we can write this as:
$$F(x)=F(x_0)+ \frac{d}{dx}F(x_0) \Delta x + O(\Delta x^2)$$
This is doable because $\Delta x$ being small means that the quadratic term will be very small and can be "ignored."
We can also do this in multiple dimensions, I will let you look up the wiki for more details but in multivariate calculus the Taylor expansion becomes:
$$F(x,y)=F(x,y)|_{(x_0,y_0)}+\frac{\partial}{\partial x}F(x,y)|_{(x_0,y_0)} \delta x+\frac{\partial}{\partial y}F(x,y)|_{(x_0,y_0)}
 \delta y+ O(\delta x^2)+O(\delta y^2)$$
This is how we get the LHS that you have written in your post. Now from there, the constant term cancels out ($L(x,\dot{x},t)$) and the second order terms ($O(\partial x^2)$) are taken to be small enough that they can be ignored. Hope this helps.
A: The $O$ is big-O notation. $O\!\left(\delta x^2\right)$ stands in place of a function whose absolute value is smaller than $c\,\delta x^2$ for some $c\gt0$.
The partial derivatives are from the linearization of the integrand:
$$
L(x+\delta x,\dot x+\delta\dot x,t)=L(x,\dot x,t)+\frac{\partial L}{\partial x}\delta x+\frac{\partial L}{\partial\dot x}\delta\dot x+O\!\left(\delta x^2\right)\left[+O\!\left(\delta\dot x^2\right)\right]
$$
Mixed partials applies to give $\delta\dot x=\frac{\mathrm{d}}{\mathrm{d}t}\delta x$ and then integration by parts allows us to convert $\delta\dot x$ to $\delta x$.
