# Weird Issue with the Chain Rule

1. Homework Statement

I'm working on a quick problem regarding a presentation that I'm giving, but I've come across an issue that I can't seem to resolve. Namely

$$\displaystyle \left. \frac{d}{dt} \right|_{t=0} f(\phi^p (t+t_0) ) = \left( \phi^p \right) ^\prime (t_0) f$$

Does anybody see how this is true?

3. The Attempt at a Solution

$$\displaystyle \left. \frac{d}{dt} \right|_{t=0} f(\phi^p (t+t_0) ) = f^\prime(\phi^p(t_0)) \left(\phi^p \right)^\prime (t_0)$$

All we know about f is that it is a smooth function and $t_0$ was arbitrarily chosen, so I'm not seeing where we make the jump. (Note: $\phi^p(t)$ is a dynamical system flow on a smooth manifold, but I don't see how that should help)

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Dick
Homework Helper
Is f some form of exponential, so f'(x)=f(x)?

All we know is that f is a smooth function on the manifold.

quasar987
Homework Helper
Gold Member
That first equation looks incomplete to me. There is no argument to f.

f is defined as a function on the manifold so the argument is generally omitted; otherwise, we would need to introduce an atlas on the manifold and define a local coordinate system using a coordinate chart. I don't think it's really necessary to the argument. Furthermore, $(\phi^p)^\prime$ is a locally defined vector field, and $\phi^p$ being a flow would imply that vector field is smooth. Thus it acts as an operator on f, in which case the argument for f is again not necessary.

I think the answer was staring me in the face the whole time.

I probably should have defined flows, it would've made this a bit more obvious.

The flow maps the Cartesian product of a monoid and a manifold to a manifold. But $\frac{d}{dt}$ is a derivative that holds only with respect to the monoid. The way that that is written isn't implying that $\phi^$ is in the argument of f, it's implying that this is indeed the product of two smooth functions, one of which is completely independent of t!

I think that must be the answer, but I really blame horrible and ambiguous notation...