Derive a new equation from $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$\implies m^2=\frac{m_0^2}{1-\frac{v^2}{c^2}}$$
$$\implies m^2c^2-m^2v^2=m_0^2c^2$$
Differentiating the equation,
$$2m \;dm\;c^2-2m\;dm\;v^2-2v\;dv\;m_0^2=0$$
Our book says when we differentiate $m^2c^2-m^2v^2=m_0^2c^2$. We will get the above equation. But, what I understand about Calculus. That I can't derive it anyway. Actually, what we differentiate here? $mass$ or, $velocity$?
Relativistic mass equation :
$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Above question is a constant equation which maybe found from Lorentz Transformation. In the second line I’ve just squared both side. In third line I‘ve just moved "something" right to left. Then, I differentiate. But, I can't understand how they differentiate here.

After writing the question, I was doing some the sum again.
Relativistic mass changes over time (How fast you travel through space your mass decreases). Here $m$ is relativistic mass and $m_0$ is "normal" mass.
So, I decided to differentiate mass
$$(mc)^2-(mv)^2-(m_0c)^2=0$$
We know,
$$(f(x))^n=n(f(x))^{n-1} . f' (x)$$
Then :
$$2 (mc) . c - 2 (mv) .v -$$
Then, I can't differentiate anymore. $c$ is speed of light which is constant. That $m_0$ is also constant. So, I stopped there.

Prove of $E=mc^2$,
$$F=\frac{dp}{dt}$$
$$=\frac{d}{dt} (mv)$$
$$= m \frac{dv}{dt} + v \frac{dm}{dt}$$ ----------------1
$$dW=F .dS$$
$$=>dK=F .dS$$
$$=>dK=[m \frac{dv}{dt} +v \frac{dm}{dt}] .dS$$
$$=m\frac{dS}{dt} .dv + v . \frac{dS}{dt} .dm$$
$$=mvdv+v^2dm$$ -------------------2
$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$\implies m^2=\frac{m_0^2}{1-\frac{v^2}{c^2}}$$
$$\implies m^2c^2-m^2v^2=m_0^2c^2$$
$$2m \;dm\;c^2-2m\;dm\;v^2-2v\;dv\;m_0^2=0$$
$$c^2dm=mvdv+v^2dm$$ ---------------3
$$dK=c^2dm$$
$$\int = \int c^2dm$$ (In LHS Integral starts from 0 and finishes at $k$. In RHS Integral starts from $m_0$ and ends at $m$)
$$k=c^2[m-m_0]$$
Hence,
$$Total energy = k + Rest mass energy$$
$$E=c^2[m-m_0]+m_0c^2$$
$$=mc^2-m_0c^2+m_0c^2$$
$$=mc^2$$
$$E=mc^2$$
 A: Say you are given
$$
m(t) = \frac{m_0}{\sqrt{1 - \frac{v(t)^2}{c^2}}},
$$
for some $t> 0$, with constants $m_0$ and $c$. Then, differentiating with respect to time, we get that
$$
\begin{align}
\frac{dm(t)}{dt} = \dot{m}(t) &= m_0 \frac{d}{dt}\left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{1}{2}} \\
&= \frac{-m_0}{2c^2} \frac{d}{dt}\left(v(t)^2\right) \left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{3}{2}} \\
&= -\frac{m_0 v(t) \dot{v}(t)}{c^2} \left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{3}{2}} \tag{1}
\end{align}
$$
In Equation (1), using the fact that
$$
\left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{3}{2}} = \left(\frac{m(t)}{m_0}\right)^3,
$$
we find that
$$
\dot{m}(t) = -\frac{m_0v(t)\dot{v}(t)}{c^2}\left(\frac{m(t)}{m_0}\right)^3
$$
Simplifying further, we obtain that
$$
\dot{m}(t) m_0^2 c^2 = \dot{v}(t) v(t) m(t)^3
$$
Instead, in Equation (1), using the fact that
$$
\left(1 - \frac{v(t)^2}{c^2}\right)^{-\frac{1}{2}} = \left(\frac{m(t)}{m_0}\right),
$$
we can instead obtain that
$$
\begin{align}
\dot{m}(t) = - \frac{m_0v(t)\dot{v}(t)}{c^2}\left(1 - \frac{v(t)^2}{c^2}\right)^{-1}\left(\frac{m(t)}{m_0}\right)
\end{align}.
$$
Simplifying this then gives us
$$
\dot{m}(t)(c^2 - v(t)^2) = -v(t)\dot{v}(t)m(t)
$$
That is, putting everything to one side
$$
\dot{m}(t) c^2 + \dot{v}(t) v(t) m(t) - \dot{m}(t) v(t)^2 = 0,
$$
which is similar to what you require.
A: What your question had was a differential equation for energy in terms of two variables, $m$ and $v$. But we could write it in terms of one variable if we knew the Jacobian, which is a fancy word for the scale factor when you convert between $v$ and $m$ variables. This is similar to the manipulation you have seen before in implicit differentiation
$$x^2+y^2=r^2 \implies 2x + 2y\cdot \frac{dy}{dx} = 0$$
but this manipulation is different because it doesn't rely on a parametric variable (one variable inside both - in the above case $x = x(x)$ and $y = y(x)$). It's just like the manipulations you've see for $u$ substitution
$$u = \tan x \implies du = \sec^2x\:dx $$
$$\tan^{-1}u = x \implies \frac{du}{1+u^2} = dx$$
and so on. In physics we can always interpret the Jacobian as the ratio of the two changes i.e.
$$\frac{dx}{du} = \cos^2 x = \frac{1}{1+u^2}$$
What the manipulation does is answer the following question: if we have a little change in each of the variables, how does the little change in one relate to the little change in the other
$$f = f(x,y) \implies df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$
In this specific case, since we know $f(m,v)$ is a constant we have that
$$df = 0 = \frac{\partial f}{\partial m}dm + \frac{\partial f}{\partial v}dv \implies \frac{dv}{dm} = -\frac{\frac{\partial f}{\partial m}}{\frac{\partial f}{\partial v}}$$
which is actually a famous theorem called the implicit function theorem. With this Jacobian in hand, we are free to replace $dv$, the change in velocity, in the following equation with the related change in mass, $dm$ by
$$dv = \frac{dv}{dm}dm$$
or, the Jacobian (scale factor) times the change in mass.
Again, this doesn't require $m$ nor $v$ to be functions of a parametric time variable. This is important for relativity because time is not suitable to be a universal parametric variable. Both time and spatial coordinates are parametrized in terms of other variables, such as the arclength of a worldline (proper time) or just an arbitrary variable with no physical meaning.
