<s>[INST] Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS? 
$$\frac{d}{dt}\left(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{m \vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$
What is the differentiation rule used?
 [/INST] Leave $m$ out and start with,
$$\frac{\vec{u}\cdot\vec{u}} {\sqrt{1-\frac{u^2}{c^2}}}=\frac{u^2} {\sqrt{1-\frac{u^2}{c^2}}}$$
Take derivatives on both sides,
$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} +\frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}
=\left[ \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}  + \frac 12 \frac{\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \right]\frac{du^2}{dt}$$
Since $ du^2/dt = 2\vec{u}\cdot d\vec{u}/dt$, rearrange the RHS as,
$$ RHS = \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}+\left[ \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}  + \frac{\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \right]\vec{u}\cdot \frac{d\vec{u}}{st}$$
$$= \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}+ \frac{\vec{u}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}}\cdot \frac{d\vec{u}}{st}$$
Compare with the LHS to cancel the first term. Thus,
$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{\vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$
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