Derivative of the transformed explanatory variable I have an explanatory variable that is transformed as suggested in footnote 25 of the article as follows (the explanatory variable is continuous and can take negative, zero or positive values)
\begin{equation}
y=\text{sign}(x)\,\log{(|x|+1)}
\end{equation}
where $\text{sign}(x)$ takes a value of $1$ if $x>0$, $0$ if $x=0$, and $-1$ if $x<0$. Let's suppose my dependent variable is $y$ (not transformed). Now I need to take the derivative of $y$ with respect to $x$ to find out the marginal effect and elasticity. 
I did this as follows: 
\begin{equation}
\frac{\partial y}{\partial x}=\text{sign}(x)\,\frac {1}{(|x|+1)}\,\frac{\partial|x|}{\partial x}
\end{equation}
As far as I understand, $\frac{\partial|x|}{\partial x} =1$ if $x>0$ and $\frac{\partial|x|}{\partial x} =-1$ if $x<0$, and is not defined for $x=0$. My question is how I compute the marginal effect for the observation with the value $x=0$ with this transformation. 
 A: While it looks like its the case that as written the derivative isn't defined there, let's look more carefully. Note that, apart from at zero, 
$\text{sign}(x)$ and $\frac{\partial|x|}{\partial x}$
are both the same quantity and the inverse of each other.
That is, except at 0, 
\begin{equation}
\frac{\partial y}{\partial x}=\frac {1}{(|x|+1)}
\end{equation}

And then the question is, can we reasonably replace the missing point with its limiting value (the value 1 at x=0)
Let's look at the original function $z$:

And we then realize that the function is not only continuous but actually smooth right through 0.
That is, the way we've written the function down and then performed our manipulations has led us to fool ourselves about the derivative being a problem. Clearly the correct value for that derivative at zero is actually 1, and we should fill the point in with its limit without being overly concerned that we did something nefarious.
The derivative of the function can simply be taken as 
\begin{equation}
\frac{\partial y}{\partial x}=\frac {1}{(|x|+1)}
\end{equation}
including at zero. (If you're in doubt, start with that derivative, and integrate it to produce the original function (using that y=0 at x=0 fixes the "+C"), and then work from there.)
(Indeed, if you go back to brass tacks and do it like when we were very first learning how to take derivatives, as $\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$, there's no difficulty.)
