Relativistic Rapidity question

[SilasG]

The question is as follows:
We can define the rapidity, y, of a particle with respect to the x axis
y≡tanh^-1β_x. Show that under a Lorentz transformation by rapidity y_B
y'=y-y_B

The Attempt at a Solution

I started by working backwards (sorry if the LaTeX does not work
$$ y'=y-y_B=\frac{1}{2}(\ln(\frac{1+\beta_x}{1-\beta_x})-\ln(\frac{1+\beta_B}{1-\beta_B}))=\frac{1}{2} \ln (\frac{1+\beta_x-\beta_B-\beta_x \beta_B}{1+\beta_B-\beta_x-\beta_B \beta_x})\\ $$
But now I am stuck, what is the transform in this case?

 

[Orodruin]

How do velocities transform under Lorentz transformations? What does this tell you about the relation between $\beta_B$, $\beta_x$ and the sought velocity $\beta'$?

sorry if the LaTeX does not work

You need to use two $ instead of one in order to get in and out of LaTeX mode. I fixed this in your original post, but it is a good thing to know.