# Confusion with the variational operator $\delta$ and finding variations

I have recently started studying String Theory and this notion of variations has come up. Suppose that we have a Lagrangian $$L$$ such that the action of this Lagrangian is just $$S=\int dt L.$$ The variation of our action $$\delta S$$ is just $$\delta S=\int dt \delta L.$$ I have read on other posts that the variation is defined to be $$\delta f=\sum_i \frac{\partial f}{\partial x^i}\delta x^i,$$ which seems like an easy enough definition. Having this definition in mind, I proceded to come across an action $$S=\int dt \frac12m \dot X^2-V(X(t))$$ which implies our Lagrangian is $$L(t,X, \dot X)$$ which makes our first varition follow as $$\delta S=m \int dt\frac12(2 \dot X)\delta \dot X-\int dt \frac{\partial V}{\partial X} \delta X$$ $$=-\int dt \left(m \ddot X+\frac{\partial V}{\partial X}\right)\delta X.$$ My question is, did that variation of the action follow the definition listed above? That is $$\delta S=\int dt\frac{\partial L}{\partial t} \delta t+\frac{\partial L}{\partial X} \delta X+\frac{\partial L}{\partial \dot X}\delta \dot X,$$ where the $$\frac{\partial L}{\partial t} \delta t$$ term vanishes because there is no $$t$$ variable.

I guess you meant $$\delta f = \sum_i \frac{\partial f}{\partial x_i} \delta x_i$$ on your third equation. Also you've implicity fixed inital $$t_0$$ and final $$t_1$$, so that your action integral really is $$S = \int_{t_0}^{t_1} dt L$$ and therefore, since the limits are fixed, variation "commutes" with integration:$$\delta \int_{t_0}^{t_1} dt L = \int_{t_0}^{t_1} dt \delta L$$ (you can check out some problems where the end intervals are not fixed for variational problems and some extra stuff is needed - see Elsgolc).
If the Lagrangian only depends on time through $$X$$ or $$\dot{X}$$, then we say that the Lagrangian has implicit but not explicit time dependence. So in your example, we would write $$$$L(X, \dot{X}) = \frac{1}{2} m \dot{X}^2 - V(X)$$$$ even though $$X=X(t)$$ depends on time. Given this Lagrangian, your variation is completely fine; the equations of motion are $$$$m \ddot{X} + V'(X) =0$$$$
In order to have explicit time dependence, you would need to have time appear explicitly, not simply through $$X$$ or $$\dot{X}$$. For example: $$$$L(X, \dot{X}, t) = \frac{1}{2} m \dot{X}^2 - V(X) + J(t) X$$$$ for some function of time $$J(t)$$. In this case, the equation of motion would be $$$$m \ddot{X} + V'(X) = J(t)$$$$