<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] Using the product rule reveals
$$\begin{align}
\frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=m\vec u \frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)+\frac{1}{\sqrt{1-u^2/c^2}}\frac{d}{dt}\left(m\vec u\right)\\\\
&=m\vec u\left(-\frac12(1-u^2/c^2)^{-3/2}\left(\frac{-2u}{c^2}\right)\,\frac{du}{dt}\right)\\&+(1-u^2/c^2)^{-3/2}(1-u^2/c^2)\frac{d}{dt}\left(m\vec u\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(\vec uu\frac{du}{dt}+(c^2-u^2)\frac{d\vec u}{dt}\right)\tag1
\end{align}$$
Forming the inner product of the right-hand side of $(1)$ with $\vec u$, we find that
$$\begin{align}
\frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^3\frac{du}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \frac12\frac{du^2}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \frac12 \frac{d(\vec u\cdot \vec u)}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \vec u\cdot  \frac{d\vec u}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=m(1-u^2/c^2)^{-3/2}\vec u\cdot \frac{d\vec u}{dt}
\end{align}$$
as was to be shown!
</s>