# Thermodynamics derivatives: Does $(\partial A/\partial B) = (\partial B/\partial A)^{-1}$?

If I need to find, for instance: $$\mathcal{T} := \left(\frac{\partial U}{\partial S}\right)_{V, N}$$ knowing: $$\left(\frac{\partial S}{\partial U}\right)_{V, N} = \frac{1}{T}$$ Can I say that $$\mathcal{T} = (1/T)^{-1} = T$$ ? In general terms, let $$A$$ and $$B$$ are two thermodynamics values, is it true that $$(\partial A/\partial B) = (\partial B/\partial A)^{-1}$$ ?

• The question is better suited for Math.SE, where it already exists. Commented Jun 6, 2022 at 13:45
• There are singularities like near 0 K or the liquid critical point, where some partial derivatives converge to 0 or $\pm \infty$ Commented Jun 6, 2022 at 14:01
• Usually monotonous , but there are exceptions, like local minima for water heat capacity near 36 °C, or for molar volume near 4 °C. Or maxima of electric conductivity of concentrated electrolytes. Commented Jun 6, 2022 at 14:06
• Additionally, such partial derivatives may be undefined at some points, typically at phase changes. Commented Jun 6, 2022 at 14:26
• @andselisk This question is about thermodynamics, and belongs here. The Math.SE answer says it's not generally the case that, for a function of more than one variable (e.g., z(x,y)), that (∂z/∂x)_y = 1/(∂x/∂z)_y. However, in thermodynamics, we're generally dealing with equilibrium systems, which means we are operating under the constraint that dz = 0. And under that constraint, we can derive the cyclic rule and the relationship that (∂z/∂x)_y = 1/(∂x/∂z)_y. So yes, this formula should be used (and often needs to be used) in thermodynamic derivations. Just watch out for singularities. Commented Jun 6, 2022 at 22:46