
Differentiation
I am trying to prove that the group velocity is equal to the particle velocity by relativistic purpose
The relativistic energy is given by the relation
$\displaystyle E^2 = p^2c^2 + m_{0}^2c^4 $
But I don't know how to differentiate it with respect to momentum "p"

Well, first you get E by itself:
$\displaystyle E = \sqrt{p^2c^2 + m_0^2c^4}$
Now, we take the derivative with respect to $\displaystyle p$ on both sides:
$\displaystyle \frac{dE}{dp} = \frac{d}{dp} \sqrt{p^2c^2 + m_0^2c^4}$
First we use the chain rule setting $\displaystyle u = p^2c^2 + m_0^2c^4$
$\displaystyle \frac{dE}{dp} = \frac{1}{2\sqrt{p^2c^2 + m_0^2c^4}}\frac{d}{dp}(p^2c^2 + m_0^2c^4)$
Now, you differentiate what is in the parentheses:
$\displaystyle \frac{dE}{dp} = \frac{1}{2\sqrt{p^2c^2 + m_0^2c^4}}(2pc^2)$
The two's cancel, and a c factors out of the radical in the denominator, which simplifies the derivative to:
$\displaystyle \frac{dE}{dp} = \frac{pc}{\sqrt{p^2 + m_0^2c^2}}$
And that's the derivative, maybe you can continue with this.