The rules of partial differentiation

$\bullet$ Identify the independent variables, eg and .
$\bullet$ If , the partial derivative of with respect to is obtained by holding constant; it is written

$\begin{displaymath} \left({\partial w\over\partial u}\right)_{\!\scriptstyle v} \end{displaymath}$

$\bullet$ It follows that

$\begin{displaymath} \left({\partial v\over\partial u}\right)_{\!\scriptstyle v} ... ...=\left({\partial v\over\partial v}\right)_{\!\scriptstyle u}=1 \end{displaymath}$

$\bullet$ The order of differentiation doesn't matter:

$\begin{displaymath} \left({\partial\over\partial v}\left({\partial w\over\partia... ...partial v}\right)_{\!\scriptstyle u}\right)_{\!\scriptstyle v} \end{displaymath}$

$\bullet$ The change in

as a result of changes in

and

$\begin{displaymath} {\rm d}w=\left({\partial w\over\partial u}\right)_{\!\script... ...{\partial w\over\partial v}\right)_{\!\scriptstyle u} {\rm d}v \end{displaymath}$

(2.1)

$\bullet$ We could take

and

to be the independent variables, with

. Now the partial derivatives are

$\begin{displaymath} \left({\partial u\over\partial v}\right)_{\!\scriptstyle w}\... ...uad\left({\partial u\over\partial w}\right)_{\!\scriptstyle v} \end{displaymath}$

Note the first is no longer required to be zero - it's

, not

, that is held constant. In this case,

$\begin{displaymath} {\rm d}u=\left({\partial u\over\partial v}\right)_{\!\script... ...{\partial u\over\partial w}\right)_{\!\scriptstyle v} {\rm d}w \end{displaymath}$

(2.2)

$\bullet$ By comparing (2.1) and (2.2) with ${\rm d}v=0$ , we see that

$\begin{displaymath} \left({\partial w\over\partial u}\right)_{\!\scriptstyle v}=... ...aystyle{\partial u\over\partial w}}\right)_{\!\scriptstyle v}} \end{displaymath}$

(2.3)

$\bullet$ From (2.1) we have

$\begin{displaymath} \mbox{\parbox{10cm}{ \begin{eqnarray*} \left({\partial w\ove... ...\right)_{\!\scriptstyle u} \!\!\!&=&\!\!\!-1 \end{eqnarray*}}} \end{displaymath}$

(2.4)

In the second line, ``dividing ${\rm d}u$ by ${\rm d}v$ '' gave $\left({\partial u\over\partial v}\right)_{w}$ , not ${{\rm d}u\over {\rm d}v}$ , because the first line was only true for constant .
$\bullet$ Rearranging (2.4) also gives

$\begin{displaymath} \left({\partial u\over\partial v}\right)_{\!\scriptstyle w} ... ...aystyle{\partial v\over\partial w}}\right)_{\!\scriptstyle u}} \end{displaymath}$

(2.5)

The minus sign in these is counter-intuitive.

Equations (2.3), (2.4) and (2.5) are our main results, and may be new to you.

References

Bowley and Sánchez E.1

Previous: 2.12 Maxwell's Relations

Judith McGovern 2004-03-17