I was reading the following
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When I taylor expand as the author says, I get
$$ \int_{}^{} \delta x\frac{\partial f}{\partial x} +\delta \dot{x}\frac{\partial f}{\partial \dot{x} } + \frac{1}{2}\left(\delta x^{2}\frac{\partial ^{2}f}{\partial x^{2 }}+2\delta x\delta \dot{x} \frac{\partial ^{2}f}{\partial x \partial \dot{x}} +(\delta \dot{x})^{2}\frac{\partial ^{2}f}{\partial \dot{x}^{2}} \right)+ \dots - f(x,\dot{x},t )dt. $$ I do not understand how the author gets the $O(\delta ^2)$ term.
Question: How does the author get the $O (\delta ^2)$ term?
The key thing to resolving this is that you should write $\delta{\dot{x}}$ instead of your $\delta{y}$, and then relate $\delta{\dot{x}}$ to $\delta{x}$ via an integration by parts. This is because your functional depends only on $x$, hence a variation of $x$ induces a variation in $\dot{x}$; so their variations are not independent. Ill work with the cross term and leave the other parts to you.
So you have $$\int...+\bigg(\delta{x}\delta{\dot{x}} \dfrac{\partial^2 f}{\partial x \partial \dot{x}}\bigg)+...$$
This is the correct way to write the cross term; for some reason you also have an extra $\partial$ sitting there; be sure to check this is the correct way to write this; again this is just taylor expansion (at least if you are working in a physics setting; it needs a bit more mathematical care if you're doing it fully rigorously).
The above can be rewritten as:
$$\int...+\bigg(\delta{x} \dfrac{d(\delta{x})}{dt} \dfrac{\partial^2 f}{\partial x \partial \dot{x}}\bigg)+...$$
$$=\int...+\bigg(\dfrac{1}{2} \dfrac{d((\delta{x})^2)}{dt} \dfrac{\partial^2 f}{\partial x \partial \dot{x}}\bigg)+...$$
Hence if you integrate by parts:
$$=\int...+\bigg(-\dfrac{1}{2} \delta{x}^2 \dfrac{d}{dt} \Big(\dfrac{\partial^2 f}{\partial x \partial \dot{x}}\Big)\bigg)+...$$
Which is the desired $O(\delta{x}^2)$ term.