Q:

Question concerning the Feynman Lectures of Physics

I am reading the Feynman lectures and at this point http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S3 it says as follows: 
The time derivate of the potential energy is 
$\begin{equation}
\dfrac{d}{dt}\sum\limits_{pairs}-\frac{Gm_{i}m_{j}}{r_{ij}} = \sum\limits_{pairs} \left( +\frac{Gm_{i}m_{j}}{r^2_{ij}} \right) \left( \dfrac{dr_{ij}}{dt} \right)
\end{equation}$
But
$\begin{equation}
r_{ij}=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2},
\end{equation}$
so that
$\begin{equation}
\begin{split}
\frac{dr_{ij}}{dt} = \frac{1}{2r_{ij}} \biggl[
  & 2 \left( x_{i}-x_{j} \right) \left( \dfrac{dx_{i}}{dt} - \dfrac{dx_{j}}{dt} \right)  \\
  +& 2 \left( y_{i}-y_{j} \right) \left( \dfrac{dy_{i}}{dt} - \dfrac{dy_{j}}{dt} \right)  \\
  +& 2 \left( z_{i}-z_{j} \right) \left( \dfrac{dz_{i}}{dt} - \dfrac{dz_{j}}{dt} \right) \biggr]
\end{split}
\end{equation}$
=========================================
What happens after "so that"? I don't understand what is being done mathematically?

A:

It's the chain rule. We have
$$ r_{ij} = \sqrt{w} $$
where I've defined $w$ as all those terms underneath the square root. So
$$ \frac{d r_{ij}}{dt} =\frac{d r_{ij}}{dw} \frac{dw}{dt} = \frac{1}{2 \sqrt{w}} \frac{dw}{dt} = \frac{1}{2 r_{ij}} \frac{dw}{dt} \,, $$ 
where to go to the second expression I've used the chain rule (the rest is just manipulation). To calculate $dw/dt$, you need to apply the chain rule again. Hopefully you can see how this will work. If not, consider each term under the square root seperately, and make substitutions such as $u = x_i - x_j$.

