Newton’s law states that $F=\frac{dp}{dt}$, where $p$ is the momentum of a body. In Newtonian physics, if the body has constant mass $m$, its momentum is $mv$, and Newton’s law becomes the familiar $F = m\frac{dv}{dt} = ma$. However, in special relativity, the momentum of a body is given by $$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $m_0$ is the rest mass of the body (which is constant), $v$ is its velocity, and $c$ is the speed of light (which is also constant). Show that $$F=\frac{m_0a}{{(1-{\frac{v^2}{c^2}})}^{\frac{3}{2}}}$$
So I'm having trouble with this question, since I'm suppose to only be using the provided information (and the fact that $v$, velocity, is the derivative of a position function with respect to time, a.k.a. $v=\frac{dx }{dt }$). However, whenever I try to find the derivative, with respect to time, which is essentially the what the question is asking for, I keep ending up with something like $$\frac{dp}{dt}=\frac{(m_0a)[(1-{\frac{v^2}{c^2}})+v^2]}{{(1-{\frac{v^2}{c^2}})}^{\frac{3}{2}}}$$ after having applied the quotient and chain rules! I'm not sure if I'm doing anything wrong; any help would be great. Thanks in advance! :)
I also do the derivative, having a different answer $$\frac{dp}{dt}=\frac{dp}{dv}\frac{dv}{dt}=\frac{m}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}a$$ Hope this will help