Differentiation of function There is a passage in a physics textbook I don't quite follow.  Since my question is mathematical, I've decided to post it here.  The book says:
Let $V$ be the volume of a molecule and assume $V = nr^3$.  Then, because the  incompressibility $K$ is defined as $K = -V \frac{\partial P}{\partial V}$, and pressure $P$ is defined as $P = - \frac{\partial U}{\partial V}$, one finds $K =V \frac{\partial ^2 U}{\partial V^2}$.
Using $V = nr^3$, $P = -\frac{dU}{dr} \frac{dr}{dV} = - \frac{1}{3nr^2} \frac{dU}{dr}$.  Thus the incompressibility becomes
$$K = -nr^3 \frac{dP}{dr} \frac{dr}{dV} = \frac{r}{9n} \frac{d}{dr} \left[\frac{1}{r^2} \frac{dU}{dr} \right]$$
Question 1
I don't quite see how this last expression is obtained.  I know that since $V = nr^3$ we have $\frac{dV}{dr} = 3nr^2$, so $\frac{dr}{dV} = \frac{1}{3nr^2}$.  Thus, I see how the expression for $P$ is obtained.  Also I see that for $K$ we have $K = -nr^3 \frac{d}{dr} \left[ -\frac{1}{3nr^2} \frac{dU}{dr} \right] \frac{dr}{dV}$, but I have a hard time working this out algebraically to get the expression above.  If anyone can show me the intermediate steps here, then I would be very grateful.

The book further states:
At the equilibrium position $r = r_0$, $\frac{dU}{dr} = 0$.  Thus
$$K_0 = \frac{1}{9nr_0} \left(\frac{d^2 U}{dr^2} \right)_0$$
Question 2
This confuses me. If $\frac{dU}{dr} = 0$ and $K = \frac{r}{9n} \frac{d}{dr} \left[\frac{1}{r^2} \frac{dU}{dr} \right]$, the shouldn't we get simply $K = 0$?
Any help on any of these questions will really be appreciated!
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\large\bf\mbox{Question 1}:}$
\begin{align}
\partiald{}{r}&=\partiald{V}{r}\,\partiald{}{V} = 3nr^{2}\,\partiald{}{V}
\ \imp\ \partiald{}{V} = {1 \over 3n}\,{1 \over r^{2}}\,\partiald{}{r}
\ \imp\ \partiald[2]{}{V}={1 \over 9n^{2}r^{2}}\,\partiald{}{r}\pars{{1 \over r^{2}}\,\partiald{}{r}}
\end{align}

\begin{align}
K&= V\,\partiald[2]{U}{V}\ =\
\overbrace{nr^{3}}^{\ds{=\ V\quad}}\ \overbrace{%
{1 \over 9n^{2}r^{2}}\,\partiald{}{r}\pars{{1 \over r^{2}}\,\partiald{U}{r}}}
^{\ds{=\ \partiald[2]{U}{V}}}\ =\
{r \over 9n}\,\,\partiald{}{r}\pars{{1 \over r^{2}}\,\partiald{U}{r}}
\\[3mm]&={r \over 9n}\braces{%
\partiald{\pars{1/r^{2}}}{r}\,\partiald{U}{r}
+{1 \over r^{2}}\,\partiald{}{r}\bracks{\partiald{U}{r}}}
=-\,{2 \over 9nr^{2}}\,\partiald{U}{r} + {1 \over 9nr}\,\partiald[2]{U}{r}
\end{align}

$\ds{\large\bf\mbox{Question 2}}$:
$\ds{\pars{\partiald[2]{U}{r}}_{0}}$ is a 'short notation': The second derivative is evaluated at the point where the first derivative is zero. It doesn't need to be zero. For example:
$$
\fermi\pars{x}\equiv x^{2} - 2x\,,\qquad
\fermi'\pars{1} = 0\,,\qquad\fermi''\pars{1} = 2 \not= 0
$$
A: 1.
$$
K=-V\frac{\partial P}{\partial V}=-V\frac{\partial P}{\partial r}\frac{\partial r}{\partial V}
=-nr^3\frac{\partial P}{\partial r}\frac{\partial r}{\partial V}
=-nr^3\frac{\partial P}{\partial r}\frac{1}{3nr^2}\\
=-nr^3\frac{1}{3nr^2}\frac{\partial}{\partial r}\left[-\frac1{3nr^2}\frac{\partial U}{\partial r}\right] 
=\frac{r}{9n}\frac{\partial}{\partial r}\left[\frac1{r^2}\frac{\partial U}{\partial r}\right]
$$
2.
$$
\frac{\partial}{\partial r}\left[\frac1{r^2}\frac{\partial U}{\partial r}\right]=-\frac2{r^3}\frac{\partial U}{\partial r}+\frac1{r^2}\frac{\partial^2 U}{\partial r^2}
$$
