I have two functions $u$ and $\phi$ given. I am not sure what they depend on, but I think that it is a common variable $\tau$. So $u(\tau)$ and $\phi(\tau)$. Then $\dot u$ is the derivative of $u$ with respect to $\tau$.

The derivation of a problem works if $$\frac{\dot u}{\dot \phi} = \frac{\mathrm du}{\mathrm d\phi}.$$

Is that legitimate? Within thermodynamics, where an equation of state like $f(u, \phi) = 0$ holds, I learned that I the reciprocal of a derivative is the derivative the other way around, allowing to “cancel chain rules”. I am not sure whether this would hold here as well.

This is probably just a duplicate of how to calculate $\frac{d\dot{x}}{dx}$? I can just cancel the $\mathrm d\tau$?


Formally, yes, you can just cancel the $d\tau$.

I think one can be more precise, although the notation is a bit awkward: if $\frac{d\phi}{d\tau}$ is nonzero, then the inverse function theorem says we can view $\tau$ as a function of $\phi$, $\tau(\phi)$, and $\frac{d\tau}{d\phi} = \frac{1}{\frac{d\phi}{d\tau}}$. Then we can view $u$ as a function of $\phi$: $u = u(\tau) = u(\tau(\phi))$. By the chain rule, $\frac{du}{d\phi} = \frac{du}{d\tau} \frac{d\tau}{d\phi} = \frac{du}{d\tau} \frac{1}{\frac{d\phi}{d\tau}}$.

  • $\begingroup$ Ah, the inverse function theorem holds in general if it is invertible. Good to know! $\endgroup$ – Martin Ueding May 30 '14 at 11:33

$$\frac{\dot u}{\dot \phi}=\frac{\left[\frac{du}{d\tau}\right]}{\left[\frac{d\phi}{d\tau}\right]}=\underbrace{\frac{du}{d\tau}\times \frac{d\tau}{d\phi}=\frac{du}{d\phi}}_{\text{by the chain rule}},$$

so you are correct (provided $\frac{d\phi}{dt}\neq 0$).

Note that $\frac{du}{d\tau}$ is not a fraction, though, for all intents and purposes (in terms of "cancelling" the $d\tau$ terms), one can treat it as a fraction in this case.


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