What is the justification for the variation of the Lagrangian in an action?

[Identity]

Given an action:

S = \int L(q,\dot{q},t) \,dt

The variation is:

\delta S = \int \left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta\dot{q}\right)\,dt

I'm guessing this is some type of chain rule, but I haven't been able to derive it... how is it justified?

 

[WannabeNewton]

Let x_{l}(t) be the path of least action. Consider the path x_{l}(t) + \zeta (t) where \zeta (t) << 1. So, S(x_{l}(t) + \zeta (t)) = \int_{t_{i}}^{t_{f}}L(x_{l}(t) + \zeta (t); \dot{x_{l}}(t) + \dot{\zeta }(t))dt After a taylor expansion to first order in \boldsymbol{\zeta }, S(x_{l}(t) + \zeta (t)) = \int_{t_{i}}^{t_{f}}[L(x_{l}(t), \dot{x_{l}}(t)) + \frac{\partial L}{\partial x(t)}\zeta (t) + \frac{\partial L}{\partial \dot{x}(t)}\dot{\zeta }(t)]dt (with the expansion terms along the path of least action) so the variation is given by \delta S = \int_{t_{i}}^{t_{f}}[ \frac{\partial L}{\partial x(t)}\zeta (t) + \frac{\partial L}{\partial \dot{x}(t)}\dot{\zeta }(t)]dt

 

[Identity]

Ah, thanks for that and the fast reply :)