Vertical bar notation: $\frac{d}{dt}|_{t=0}f(a+tv)=$? The definition of the directional derivative is $D_vf(a)$=$\left.\dfrac{d}{dt}\right|_{t=0}f(a+tv)=\lim \limits_{t\to0}\dfrac{f(a+tv)-f(a)}{t}$. However, I do not understand how this bar notation is exactly expanded and what its definition is. What if we had equations like follows (even if they do not necessarily make sense):


*

*$\left.\dfrac{d}{dt}\right|_{a=0}f(a+tv)$

*$\left.\dfrac{d}{dt}\right|_{s=0}f(a+tv)$

*$\left.\dfrac{d}{dt}\right|_{t=0}f(a+v)$

*$\left.\dfrac{d}{dt}\right|_{t=0}f(a)$
What are they equivalent to by definition after expansion?
To clarify myself, is the general definition 
$\left.\dfrac{d}{dt}\right|_{s=u}f(a+rv)=\lim \limits_{s\to u}\dfrac{f(a+rv)-f(a)}{t}$?

*

*means that you derive $f(a+tv)$ w.r.t. $t$ (this gives $Df(a+tv)[v]=D_vf(a+tv))$) and then you evaluate the latter at $a=0$, which gives $Df(tv)[v]=D_vf(tv)$.

*means that you derive $f(a+tv)$ w.r.t. $t$ (this gives $D_vf(a+tv)$) and then you evaluate the latter at $s=0$, which gives $D_vf(a+tv)$ since the previous results doesn't depend on $s$.

*means that you derive $f(a+v)$ w.r.t. $t$ (this gives $0$) and then you evaluate the latter at $t=0$, which gives $0$ again.

*means that you derive $f(a)$ w.r.t. $t$ (this gives $0$) and then you evaluate the latter at $t=0$, which gives $0$ again.


So the bar notation represents the evaluation at the specified value after the differentiation has been applied. Explicitly:
$\left.\dfrac{d}{dt}\right|_{s=u}f(a+rv)=
\left.(\dfrac{d}{dt} f(a+rv))\right|_{s=u}$
