Can I find $\frac{dm}{dt}$ where $m$ is the relativistic mass of a particle?

Question

So, recently I learned the basics of Special Relativity, and I found out that the mass of a body increases with the increase in its velocity as given by the Relativistic Mass equation:
$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$
So I wanted to find $\frac{dm}{dt}$ for such cases.
Here's my method:
$F=\frac{dp}{dt}$
Also,
$F=m\frac{dv}{dt}\gamma$
where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$p=mv\gamma$
Substituting value of $p$ in $F$, we get:
$F=\frac{d(mv\gamma)}{dt}$
Breaking the differentiation and equating it with $F=m\frac{dv}{dt}\gamma$, we get:
$\frac{dm}{dt}\gamma=-\frac{d\gamma}{dt}m$
After finding $\frac{d\gamma}{dt}$, and substituting,we get:
$\frac{dm}{dt}=\frac{mv\gamma}{c^2}=\frac{p}{c^2}$
Is it correct?

Answer 1

Your assumption that $F = m \frac{dv}{dt}\gamma$ is not correct. You should instead have $$F = \frac{d}{dt} (\gamma m_0v) = v\frac{dm}{dt} + m\frac{d}{dt}(v)$$ where note $m = \gamma m_0$.

However, to find $\frac{dm}{dt}$ I would rather differentiate the expression $$ m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} $$ giving $$ \frac{dm}{dt} = m_0\frac{d}{dt} \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} $$ Let me know if you need me to perform the last differentiation.