Didn't understand a step in Einstein's paper on special relativity I didn't understand a step in Einstein's paper(special relativity).
Suppose we have a function $f(x', t)$ such that:
$$\frac{1}{2}[f(0, t) + f(0, t + \frac{x'}{c - v} + \frac{x'}{c + v})] = f(x', t + \frac{x'}{c - v})$$
He then writes, "let x' be infinitesimal":
$$\frac{1}{2}(\frac{1}{c - v} + \frac{1}{c + v}) \frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{x'}} + \frac{1}{c - v} \frac{\partial{f}}{\partial{t}}$$
Can somebody explain to be how he got from the first equation to this?
 A: Disclaimer: I am a novice when it comes to infinitesimal calculus, and I am interested in the answer to this question. Below is the work that I would have shown, if this had been a question of mine. I am posting this because I do not want to post a duplicate question. Hopefully Mikhail or someone else will be able to confirm the validity of the work below, or provide a correct solution!
In my version of the paper, we have:
$$\frac{1}{2}\left[f(0,t)+f\left(0,t+\frac{x'}{c-v}+\frac{x'}{c+v}\right)\right]=f\left(x',t+\frac{x'}{c-v}\right)$$
The infinitesimal calculus tells us that $f(x + \delta x,t+\delta t) = f(x,t)+\frac{\partial f}{\partial t}\delta t + \frac{\partial f}{\partial x}\delta x+\epsilon\ \sqrt{\delta x^2 + \delta t^2}$, for infinitesimal $\delta x, \delta t,\epsilon$. So if we let $x'$ be infinitesimal, we obtain:
$$\frac{1}{2}\left[2f(0,t)+\lambda\frac{\partial f}{\partial t} x' + \epsilon\lambda x'\right]=f(0,t) + \frac{\partial f}{\partial x'}x'+\frac{1}{c-v}\frac{\partial f}{\partial t}x'+\epsilon'x'\sqrt{1+\frac{1}{(c-v)^2}}$$
$$\frac{1}{2}\left[\lambda\frac{\partial f}{\partial t} x' + \epsilon\lambda x'\right]=\frac{\partial f}{\partial x'}x'+\frac{1}{c-v}\frac{\partial f}{\partial t}x'+\epsilon'x'\sqrt{1+\frac{1}{(c-v)^2}}$$
$$\frac{1}{2}\left[\lambda\frac{\partial f}{\partial t} + \epsilon\lambda \right]=\frac{\partial f}{\partial x'}+\frac{1}{c-v}\frac{\partial f}{\partial t}+\epsilon'\sqrt{1+\frac{1}{(c-v)^2}}$$
where we have defined $\lambda = \frac{1}{c+v}+\frac{1}{c-v}$ for brevity. Dropping the infinitesimals, we find:
$$\frac{1}{2}\lambda\frac{\partial f}{\partial t} =\frac{\partial f}{\partial x'}+\frac{1}{c-v}\frac{\partial f}{\partial t}$$
