# 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!

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. $$\frac{\partial}{\partial r}\left[\frac1{r^2}\frac{\partial U}{\partial r}\right]=-\frac2{r^3}\frac{\partial U}{\partial r}+\frac1{r^2}\frac{\partial^2 U}{\partial r^2}$$

• Great! Thanks a lot. Really appreciate it! – Kristian Jul 1 '14 at 18:22


\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: $$\fermi\pars{x}\equiv x^{2} - 2x\,,\qquad \fermi'\pars{1} = 0\,,\qquad\fermi''\pars{1} = 2 \not= 0$$

• Thanks a lot! Much appreciated! Wish I could mark both answers as accepted! – Kristian Jul 1 '14 at 18:23