Lorentz Velocity Transform With Tensor Notation So I'm attempting to prove the Lorentz Velocity tranform:
$${v_x}' =\frac{v_x-u}{1-v_xu/c^2} $$
using tensor notation. In this case obviously $\beta = u/c$ and $\gamma=(1-\beta^2)^{-1/2}$. The velocity transform tensor can be represented as 
$$\Lambda = \begin{pmatrix} 
\gamma & \beta \gamma & 0 & 0 \\ 
\beta \gamma & \gamma & 0 & 0 \\ 
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 
\end{pmatrix}  $$
assuming the the boosted frame $F'$ is moving in the x-direction from frame $F$. I'm also using the following two facts:
$${\partial_v}' = {\Lambda^u}_v \partial_u \hspace{20mm} {x^v}'={(\Lambda^{-1})^v}_u x^u$$
where 
$$\partial_u = \left(\frac{1}{c}\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right)$$.
My proof begins as follows:
$${v_x}' = c{\partial_0}'{x^1}' = c({\Lambda^{u}}_0 \partial_u)({(\Lambda^{-1})^{1}}_{\sigma}x^{\sigma}) $$
I use $\sigma$ to keep the summation seperate. Now I expand:
$${v_x}'  = c({\Lambda^{0}}_0 \partial_0+{\Lambda^{1}}_0 \partial_1)({(\Lambda^{-1})^{1}}_{0}x^{0}+{(\Lambda^{-1})^{1}}_{1}x^{1}) $$
Subbing in the appropriate elements from the tensor yields:
$${v_x}' = c(\gamma \frac{1}{c}\frac{\partial}{\partial t}+\beta \gamma \frac{\partial}{\partial x})(-\beta \gamma c t+\gamma x) $$
$$=c(-\beta \gamma^2 \frac{\partial t}{\partial t}+\frac{\gamma ^2}{c}\frac{\partial x}{\partial t} - \beta^2\gamma^2c\frac{\partial t}{\partial x}+\beta \gamma^2 \frac{\partial x}{\partial x})$$
at this point I make the (perhaps incorrect) assumption that $\partial t/\partial x=1/v_x$. Canceling out the obvious terms leaves me with
$${v_x}'=\gamma^2v_x -\frac{\beta ^2 \gamma^2 c^2}{v_x} $$
which I know to be incorrect.
 A: Einstein velocity addition equation in Minkowski space
\begin{align*}  
 &\text{I) We want to show that The velocity addition equation is:} \\
  &v_g=\frac{v_1+v_2}{1+\frac{v_1\,v_2}{c^2}}\\
  &\text{ We take the coordinate  transformation between $(t'\,,x')$ and $(t\,,x)$ (c=1)}\\\\
  &\begin{bmatrix}
       t' \\
       x' \\
     \end{bmatrix}=\underbrace{\gamma(v)
  \begin{bmatrix}
       1 & -v \\
       -v\ & 1\\
     \end{bmatrix}}_{L(v)}
     \begin{bmatrix}
       t \\
       x \\
     \end{bmatrix}\\
&\text{$L(v)$ is the Lorentz transformation matrix.} \\\\
&\text{II) we take additional coordinate transformation  between  $(t"\,,x")$ and  $(t'\,,x')$ we get }
\end{align*}
\begin{align*}
\begin{bmatrix}
       t" \\
       x" \\
     \end{bmatrix}=L(v_2)
 \begin{bmatrix}
       t' \\
       x' \\
     \end{bmatrix}=    
L(v_2)\,L(v_1)\,
  \begin{bmatrix}
       t \\
       x \\
     \end{bmatrix}
&=\underbrace{\gamma(v_1)\,\gamma(v_2)\begin{bmatrix}
       1 & -v_1 \\
       -v_1\ & 1\\
     \end{bmatrix}\begin{bmatrix}
       1 & -v_2 \\
       -v_2\ & 1\\
     \end{bmatrix}}_{L(v_1\,,v_2)} \begin{bmatrix}
       t \\
       x \\
     \end{bmatrix}\\
&=\underbrace{\gamma(v_1)\,\gamma(v_2)\,(1+v_1\,v_2)}_{\gamma(v_1\,,v_2)}
\begin{bmatrix}
  1 & -\frac{v_1+v_2}{1+v_1+v_2} \\
  -\frac{v_1+v_2}{1+v_1+v_2} & 1 \\
\end{bmatrix}
\end{align*}
\begin{align*}
&\text{with:}\\\\
&\gamma(v_1\,,v_2)=\frac{1}{\sqrt{1-v_1^2}}\,\frac{1}{\sqrt{1-v_2^2}}
(1+v_1\,v_2)=\frac{1}{\sqrt{1-\left(\frac{v_1+v_2}{1+v_1\,v_2}\right)^2}}\\\
&\Rightarrow\\
&\text{The Lorentz transformation matrix between $(t"\,,x")$ and $(t,x)$ is:}\\\\
&L(v_1,v_2)=\frac{1}{\sqrt{1-v_g^2}}
\begin{bmatrix}
  1 & -v_g \\
  -v_g & 1 \\
\end{bmatrix}\quad, \text{with:}\\\\
&\boxed{v_g=\frac{v_1+v_2}{1+\frac{v_1\,v_2}{c^2}}}\quad \text{Einstein velocity addition equation in Minkowski space}
\end{align*}
A: 4-velocity is defined as
$$
 u^{\mu}=\frac{dx^\mu}{d{\tau}}
$$
$\tau$ does not change under transformation. It is the proper time.
so in your formula $$v_{x'}=c\partial_{0'}x^{1'}$$ is not the correct way. Do you see the point? 
A: I am not 100% sure you can get this result in that way, but it may be possible. The essential problem is that you are at some level using a derivative which has to do with fields, the $\nabla$ which obeys Lorentz transforms, but you are trying to use it to analyze a single world line.
Like the traditional derivation is not too hard; the Lorentz boost preserves the origin and so one can take the ratio of the 4-vector components $(ct, ~v_xt)$ to get the dimensionless velocity $v_x/c$. After the boost you have $\gamma~(ct-uv_x t/c,~v_x t-ut)$ and the ratio of those components is just the dimensionless velocity $(v_x-u)/(c-uv_x/c)$ as promised.
For your attempt probably you would come to a similar place if doing it correctly. Ignore the $y,z$ directions and define $w=ct$ and start with a field $f$ traveling at constant velocity, $$\rho(w, x)=f(x-\eta w).$$ Then you can certainly inspect the gradient to find $$
\begin{align}
\nabla_w\rho&=-\eta f'(x-\eta w),\\
\nabla_x\rho&=f'(x-\eta w).
\end{align}
$$
So we can again find the velocity $v_x=\eta~c$ as a ratio of these covector components.
If you start from there and compute the same ratio after your Lorentz transform you should get the standard result without your confusion.
