How to derive acceleration addition formula in special relativity After seeing the derivation of the velocity addition formula (wikipedia) I want to use the same approach for deriving the acceleration addition formula.
My understanding of velocity addition derivation
In two frames $S, S'$ in standard configuration ($S$ thinks $S'$ moves with speed $v$ in x-direction and $u$ is speed in x-direction).
We have
$dt' = \gamma(dt - \frac{v}{c^2}dx)$ and $dx' = \gamma(dx-v\ dt)$.
Velocity is defined as
$$
u' = \frac{dx'}{dt'} = \frac{\gamma(dx-v\ dt)}{\gamma(dt - \frac{v}{c^2}dx)}
$$
Now we divide everything with $dt$:
$$
u' = \frac{(\frac{dx}{dt}-v\ \frac{dt}{dt})}{(\frac{dt}{dt} - \frac{v}{c^2}\frac{dx}{dt})} = \frac{u - v}{1 - \frac{v}{c^2} u}
$$
which is the correct formula.

Doing the same thing for acceleration addition
We have $du' = \frac{du - v}{1 - \frac{v}{c^2} du}$
$$a' = \frac{du'}{dt'} = \frac{\frac{du - v}{1 - \frac{v}{c^2} du}}{ \gamma(dt - \frac{v}{c^2}dx)}$$
Now we divide everything with $dt$ and assuming $\frac{v}{dt}= 0$ and $\frac{1}{dt} = 0$ (which we seemed to assume in the velocity derivation) we have:
$$
a' = \frac{\frac{\frac{du}{dt} - \frac{v}{dt}}{\frac{1}{dt} - \frac{v}{c^2} \frac{du}{dt}}}{ \gamma(\frac{dt}{dt} - \frac{v}{c^2}\frac{dx}{dt})} = \frac{\frac{a}{- \frac{v}{c^2}a}}{\gamma(1- \frac{v}{c^2}u)} = \frac{1}{-\frac{v}{c^2} \gamma (1 - \frac{v}{c^2}u)}
$$
which is obviously wrong. I suspect I'm to sloppy with my differentials (how to treat them algebraically has always been a bit mysterious for me) but what exactly am I doing wrong and how would I do it if I wanted to use an approach similar approach as my velocity addition derivation?
 A: Yes, you are sloppy with differentials. 
The correct way to compute differential for function of multiple variables is:
$$
df(x_i)=\frac{\partial f(x_i)}{\partial x_j}dx_j,
$$
where you sum through the j index on the rhs.
In one dimensional case you get:
$$
df(x)=\frac{df}{dx}dx
$$
So applying this knowledge to velocity transformation, which is one dimensional function of $u$ ($u'=f(u)$):
$$
du'=\frac{d}{du}\left(\frac{u-v}{1-\frac{v}{c^2}u}\right)du=\frac{1}{\gamma^2\left(1-\frac{v}{c^2}u\right)^2}du
$$
So plugging it in for $a'$:
$$
a'=\frac{1}{\gamma^2\left(1-\frac{v}{c^2}u\right)^2}\frac{du}{\gamma(dt-\frac{v}{c}dx)}=\frac{1}{\gamma^3\left(1-\frac{v}{c^2}u\right)^2}\frac{\frac{du}{dt}}{\gamma(1-\frac{v}{c}\frac{dx}{dt})}=\frac{1}{\gamma^3\left(1-\frac{v}{c^2}u\right)^3}a
$$
On an intuitive level, differential of function can be viewed as its infinitesimal increment (and thus can be manipulated up to a certain point as ordinary number). The first thing about this is, that the infinitesimal means it is as close to zero as you can get and thus you can neglect the differential with respect to any noninfinitesimal value. So in your equation for $du'$ you can neglect $du$ terms with respect to $v$ and 1 and you will get $du'=-v$ which is nonsense. On formal mathematical level, you are adding in numerator differential $du$ and a real number $-v$ which is again nonsense, since you are adding two different mathematical entities. 
On an intuitive level again, the transformation formula for differential can be viewed in this way: $df$ is an infinitesimal increment of the function and you want to know, how does this increment depend on the increments of the function parameters. The answer is (in one dimensional case):
$$
\Delta f=f(x+\Delta x)-f(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x}\Delta x\approx \frac{df}{dx}\Delta x
$$
Where I have used $\Delta$ instead of $d$ to make explicit that we are asking about (really, really small) increment which is just a number. 
Hope this helps to clear some of your confusions.
A: Here's how I would show the velocity addition formula: 
Begin with the Lorentz Transformations:
\begin{eqnarray} 
x^\prime &=& \gamma\left(x - v t\right)\\t^\prime &=& \gamma\left(t - \frac{v}{c^2}x\right)
\end{eqnarray}
These transformations can be written in terms of differences, 
\begin{eqnarray} 
\Delta x^\prime &=& \gamma\left(\Delta x - v \Delta t\right)\\\Delta t^\prime &=& \gamma\left(\Delta t - \frac{v}{c^2}\Delta x\right)
\end{eqnarray}
Now, dividing these two equations, we get (as you showed):
\begin{equation}
\frac{\Delta x^\prime}{\Delta t^\prime} = \frac{\Delta x- v \Delta t}{\Delta t - \left(\frac{v}{c^2}\right)\Delta x}
\end{equation}
Using the fact that $$u^\prime = \frac{\Delta x^\prime}{\Delta t^\prime} \quad \text{ and }\quad u = \frac{\Delta x}{\Delta t}$$ and dividing the right hand side by $\Delta t$, we get 
$$u^\prime = \frac{u-v}{1-\frac{uv}{c^2}}.$$
Note, however, that we have never said that $v/\Delta t = 0$, or that $1/\Delta t = 0$.
Now, if you want to do this for the acceleration relation, you will need to find how $\Delta u^\prime$ is related to $\Delta u$. Of course, since these relations are not linear, it's not as easy to get them, but a quick trick is to use the rules of differentiation. Since we know from the chain-rule that $$u^\prime = \frac{u-v}{1-\frac{u v}{c^2}}, \implies \text{d} u^\prime = \frac{\text{d}(u-v)}{1-\frac{uv}{c^2}} + (u-v) \text{d}\left(\frac{1}{1 - \frac{u v}{c^2}}\right),$$
from which it's easy to show
$$\text{d}u^\prime = \text{d} u \left(\frac{1}{\left(1 - \frac{u v}{c^2}\right)} + \frac{u-v}{\left(1 - \frac{u v}{c^2}\right)^2} \left( \frac{v}{c^2} \right)\right),$$
and so $$\text{d} u^\prime = \frac{\text{d}u}{\gamma^2 \left(1 - \frac{u v}{c^2}\right)^2}.$$
From here, I'll leave it to you to divide the entire equation by $\text{d}t^\prime$ and show that 
$$a^\prime = \frac{a}{\gamma^3 \left(1 - \frac{uv}{c^2}\right)^3}.$$
Of course, the entire above analysis is only in one dimension, and the result we've found is for $a \to a_x$ and $u \to u_x$, if you wanted to find out the other components, you could use the same method, but (if my memory serves me) they're a little more complicated.
