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Am I using the triple product rule correctly?

The triple product rule states $\displaystyle\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y=-1$

Where each derivative is the partial derivative of one variable with respect to another, while holding the third (outside the brackets) fixed.

If I apply this to $xy=z$ I get:

$\displaystyle\left(\frac{1}{x}\right)\left(\frac{-z}{y^2}\right)y=-1$

Which solves to give $-z/xy=-1$

Which all seems to check out [errata fixed]

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  • $\begingroup$ Please, give a reference for the "triple product rule", and where to find definitions of exotic things like $\displaystyle\left(\frac{\delta x}{\delta y}\right)_z$. Without precise definitions, you don't even have a stick in mathematics. $\endgroup$ – Professor Vector Aug 7 '17 at 10:02
  • $\begingroup$ @ProfessorVector sorry I've added the reference now. I made the mistake of assuming this was a well known identity. $\endgroup$ – samerivertwice Aug 7 '17 at 10:54
  • $\begingroup$ Then, you didn't even quote it correctly, let alone use. And the symbols you use mean functional derivatives, normally. $\endgroup$ – Professor Vector Aug 7 '17 at 11:06
  • $\begingroup$ @ProfessorVector thanks, I've spotted my typo. How do I get the d I want? $\endgroup$ – samerivertwice Aug 7 '17 at 11:10
  • $\begingroup$ Try "\partial y" to get $\partial y$. $\endgroup$ – Professor Vector Aug 7 '17 at 11:13
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I am afraid this is physicists' notation.

You have $z=xy$. Then $x=z/y$ and $y=z/x$. So $$\left(\frac{\partial x}{\partial y}\right)_z=-\frac{z}{y^2},$$ $$\left(\frac{\partial y}{\partial z}\right)_x=\frac{1}{x}$$ and $$\left(\frac{\partial z}{\partial x}\right)_y=y.$$ Multiplying these together gives $$-\frac{yz}{xy^2}=-\frac{xy^2}{xy^2}=-1.$$

As for interpreting what the likes of $\left(\frac{\partial x}{\partial y}\right)_z$ actually mean, good luck!

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  • $\begingroup$ Ah yes, I've made a typo. Thanks. $\endgroup$ – samerivertwice Aug 7 '17 at 11:02
  • $\begingroup$ xy=z plots a 2-d surface. That derivative gives us the ratio of the change in $x$ with respect to $y$ while we move along an isobar of $z$ on the surface. $\endgroup$ – samerivertwice Aug 7 '17 at 12:08

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