You can define the "relativistic mass" as \gamma m, where m is the mass and
\gamma=\frac{1}{\sqrt{1-v^2}}
And if you do, the relativistic mass depends on speed by definition. (I'm using units such that c=1. If you want to include c explicitly, replace every velocity in the formulas I type with velocity/c).
If you don't want to define it explicitly, you still have to define something, and take that as the starting point. For example, you can define the four-momentum p' in the comoving inertial frame as mass times the four-velocity u'=(1,0,0,0). p'=mu'=(m,0,0,0). (I'm putting a prime on the four-vectors in the comoving inertial frame, and I'm dropping the primes for the same four-vectors in the other frame). In an inertial frame that moves with velocity -\vec v relative to the particle (so that the velocity of the particle is \vec v), the four-velocity is u=\gamma(1,\vec v). (You get this result by applying a Lorentz transformation to (1,0,0,0)). So the four-momentum in this frame is (E,\vec p)=p=m\gamma(1,\vec v)=(\gamma m,\gamma m \vec v). This can be interpreted as a "derivation" of relativistic mass, because it shows that \vec p=\gamma m\vec v, instead of =m\vec v as we might have expected.
My post #14 wasn't a good answer to your question. It's really an answer to a different question. A tip for next time: When you get an answer that you feel doesn't answer your question, don't ignore it. Read it, let the guy who answered know that you read it, and let him know if it was helpful or not. It's annoying when you just ask the same question five times.
