My textbook states (in a derivation of the Euler-Lagrange equation) the following which I would like to understand: \begin{equation*} S[\bar{x}+\delta x] =\int_{t_a}^{t_b} L(\dot{x}+\delta \dot{x}, x+\delta x, t) dt =\int_{t_a}^{t_b}\left[L(\dot{x}, x, t)+\delta \dot{x} \frac{\partial L}{\partial \dot{x}}+\delta x \frac{\partial L}{\partial x}\right] d t\end{equation*} How are the integrands equal to each other? Using the definition of partial derivative, I can write the following for $\dot{x}$ \begin{equation*} \frac{\partial L}{\partial \dot{x}} =\frac{L(\dot{x}+\delta \dot{x}, x, t)-L(\dot{x}, x, t)}{\delta \dot{x}} \end{equation*}
\begin{equation*} L(\dot{x}+\delta\dot{x}, x, t) =\delta \dot{x} \frac{\partial L}{\partial \dot{x}}+L(\dot{x}, x, t) \end{equation*} Similarly, for $x$: \begin{equation*} \frac{\partial L}{\partial x} =\frac{L(\dot{x}, x+\delta x, t)-L(\dot{x}, x, t)}{\delta x} \end{equation*} \begin{equation*} L(\dot{x}, x+\delta{x}, t) =\delta x \frac{\partial L}{\partial x}+L(\dot{x}, x, t) \end{equation*} If I add the two expressions, I get: \begin{equation*} \delta\dot{x}\frac{\partial L}{\partial \dot{x}}+\delta x \frac{\partial L}{\partial x}+2 L(\dot{x}, x, t)=L(\dot{x}, x+\delta x, t)+L(\dot{x}+\delta \dot{x}, x, t) \end{equation*} By Identification with the original statement: \begin{equation*} L(\dot{x}, x+\delta x, t)+L(\dot{x}+\delta \dot{x}, x, t)+L(\dot{x}, x, t)=L(\dot{x}+\delta \dot{x}, x+\delta x, t) \end{equation*} Why is this true, and generally how would you prove the statement yourself? I feel like there is a basic fact of calculus that I am forgetting. Some help would be lovely.