Why can you differentiate inside an infinitesimal generator? I was looking at this paper page 3 right under section 2: https://www.lpsm.paris/pageperso/zhan/pdffile/rwre-survey.pdf
We're interested in the process satisfying:
$$dX_V(t)=-\frac{1}{2}V'(X_V(t))dt+dB_t$$
It says if $V$ is not differentiable we can think of this as the process with generator 
$$\frac{1}{2}e^{V(x)}\frac{d}{dx}\left(e^{-V(x)}\frac{d}{dx}\right)$$
But $V$ is still being differentiated here. How does this help? 
Edit: I think the key phrase is "Feller canonical form" of generator. I am trying to understand that.
 A: As I am not an expert on the subject, this is not a complete answer (therefore does not deserve the bounty), still I hope it might help you.
If you start with an infinitesimal generator which is a (formal) second order differential operator, 
$$
a(x) \partial^2_x + b(x) \partial_x,
$$
where $a$ is the so called diffusion and $b$ is the so-called drift,
you can manipulate it in the following way
$$
a(x) \partial^2_x + b(x) \partial_x \\
=
a(x) \left(\partial_x + \frac{b(x)}{a(x)}\right) \partial_x  \\
=
a(x) \left(e^{-\int\frac{b}{a}} \partial_x e^{\int\frac{b}{a}} \right) \partial_x \\
=
\left( a(x) e^{-\int\frac{b}{a}} \partial_x \right) \left(e^{\int\frac{b}{a}} \partial_x \right) \\
=
\partial_m \partial_s
$$
where the scale function ($s$) and the (distrubition function of) speed measure ($m$) are defined as
$$
\frac{ds}{dx} = \exp\left(-\int\frac{b}{a}\right) \\
\frac{dm}{dx} = \frac{1}{a(x) s'(x)}
$$
Here, I am using the notion of derivative with respect to (the distribution function of) a measure: 
$$
\partial_m u(x) =  \lim_{h \to 0} \frac{u(x+h) - u(x)}{m(x+h) - m(x)}
$$
This is sometimes referred to as the Feller canonical form of the generator.
If you started with smooth diffusion and drift coefficients, you'll end up with smooth scale function and speed measure.
But the point here is that the scale and speed do not need to be smooth, so you can start with non-smooth $s$ and $m$ and still define a valid generator. Here I am not sure, but I think that $s$ only needs to be strictly increasing and continuous and the distribution of $m$ only stricly increasing and right-continuous. 
In the case at hand, we have that
$$
a = \frac12 \\
b = -\frac12 V'
$$
so that the scale function is $\int e^{V}$, the speed measure is $2 \int e^{-V}$
and no need to differentiate $V$
Does this help?
