# Mixing total and partial derivatives

I'm looking at a problem in kinetic theory, where the distribution function depends on two variables: time $t$ and energy $U$. $dU/dt \ne 0$, but there is a function $J(U, t)$ with $dJ/dt = 0$. While trying to analyze this, I keep coming up with these terms $$\frac{\partial}{\partial t}\left(\frac{dU}{dt}\right), \; \frac{\partial}{\partial U}\left(\frac{dU}{dt}\right)$$ My instinct says they're zero, but I'm having trouble showing it. Any help?

• Unfortunately, I think it's easy to show that your instinct is incorrect here, for example, lets just say that $U = t^2$. What does that say about the first term? – DilithiumMatrix Jan 13 '17 at 16:49
• @DilithiumMatrix Well I'm not entirely sure, because $\partial/\partial t$ is holding $U$ constant. Consider $\partial(df/dt)/\partial t$, where $f = f(U,t)$. Now $df/dt = \partial f/\partial t + (dU/dt)\partial f/\partial U$, but when you take $\partial/\partial t$ of that you get the term I have above. – eyeballfrog Jan 13 '17 at 19:06

You have $$\frac{\mathrm d J(U(t),t)}{\mathrm d t}=0.$$ The derivative is: $$\frac{\mathrm d J(U(t),t)}{\mathrm d t}=\frac{\partial J(U(t),t)}{\partial t}+\frac{\partial J(U(t),t)}{\partial U(t)} \frac{\mathrm d U(t)}{\mathrm d t}.$$
Just try it out by choosing something specific for $U(t)$ and $J(U,t)$.