Lagrangian of a free particle, from Landau mechanics page 6 Let $K$ and $K'$ be two inertial frame and $K'$ moving with an infinitesimal velocity $\epsilon$ relative to $K$, i.e. $\vec{v'} = \vec{v} + \epsilon$. Let $L$ be the Lagrangian of $K$, $L'$ be the Lagrangian of $K'$ then we have $L' = L(v'^2) = L(v^2 + 2 \vec{v}\cdot\epsilon+ \epsilon^2)$ We can use Taylor series expansion and neglect second order terms: $$L(v'^2) = L(v^2) + 2\frac{\partial L}{\partial v^2}\vec{v}\cdot\epsilon$$ Landau then proceeds to argue that when $2\frac{\partial L}{\partial v^2}\vec{v}\cdot\epsilon$ is a linear function of $\vec{v}$, $L(v'^2) - L(v^2)$ is a total time derivative. Therefore he deduced that $L = \frac{m}{2}v^2$ My question is: Why $2\frac{\partial L}{\partial v^2}\vec{v}\cdot\epsilon$ is a linear function of $\vec{v}$? Why it is a total derivative of time of function $f(x,t)$?
Trouble understanding Landau & Lifshitz writing about Lagrangians and Galilean Relativity We have two inertial coordinate systems, $K'$ and $K$. $K$ is moving with infinitesimal velocity ${\epsilon}$ relative to $K'$. Using Galilean relativity we can transform this into $v'=v+{\epsilon}$. Here Lagrangian is only a function of speed $L=L(|v|^2)$. $L'$ may differ from $L$ only by a total derivative of function of coordinates and time, i.e. $L'=L+\frac{d}{dt}f(q,t)$  $"$We have $L'=L(|v|^2+2{v}·{\epsilon}+{\epsilon}^2)$. [1. ] Expanding this expression in powers of ${\epsilon}$ and neglecting terms above the first order, we obtain $L(|v'|^2)=L(|v|^2)+\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ [2.] The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $v$ $"$  I understand nothing in the quoted part. How do we get [2.]? Why does $\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ being a linear function of velocity imply that it's a total time derivative?
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