Differentiating curves given by curves in a Lie group acting on a point This is a problem I keep running into and I feel it is likely a really basic fact that I am forgetting/missing but I haven't been able to see a way through. Suppose I have a curve in a Lie group: $\varphi:I\to G$ and I let that act on a point $v$ in some representation $V$ of $G$ to make a curve there.
$$\sigma := \varphi \cdot v : I \to V.$$

How can I write the derivative $\sigma'$ in terms of $v$ and $\varphi$?

For simplicity I am happy if that representation is the adjoint one $V = \mathfrak{g}$ and conversely for more complexity I would love an answer for when $v$ is replaced by a map $I \to V$.
One guess might be to use the logarithmic derivative $\varphi^{-1} \cdot \varphi'$ in some way so we can get a curve in the Lie algebra to act on $v$:
$$ \varphi^{-1}(\varphi \cdot v)' = (\varphi^{-1} \cdot \varphi')\cdot v.$$
Both sides here are curves in $V$ but I can't see why such a thing would be true. Is there some approachable way of computing $(\varphi \cdot v)'$ here or is this a foolish endeavour?
 A: I think this should work out well if you use the appropriate logarithmic derivative (which I think is the right logarithmic derivative in this case). Let $\ell:G\times V\to V$ be the smooth map defining the representation of $G$ viewed as a left action. Fixing one of the entries you get the partial maps $\ell_g:V\to V$ for $g\in G$ and $\ell^v:G\to V$. These are smooth and $\ell_g$ is linear. Now taking curves $\phi:I\to G$ and $v:I\to V$, you want to compute derivatives of $\phi(t)\cdot v(t)=\ell(\phi(t),v(t))$. For the derivative of the curve, you simply get $(\phi'(t),v'(t))$ so overall, one obtains
$$
\frac{d}{dt}\ell(\phi(t),v(t))=T_{v(t)}\ell_{\phi(t)}(v'(t))+T_{\phi(t)}\ell^{v(t)}(\phi'(t)). 
$$
The first term poses no problems, since $\ell_{\phi(t)}$ is linear, so this simply gives $\ell_{\phi(t)}(v'(t))=\phi(t)\cdot v'(t)$. For the second term, you can proceed as follows: By definition $\ell(hg,w)=h\cdot (g\cdot w)$ so denoting by $\rho^g$ right translation by $g$, you get $\ell^w\circ\rho^g=\ell^{g\cdot w}$. Second, the definition of the infintesimal representation of the Lie algebra $\mathfrak g$ of $G$ on $V$ says that the map $T_e\ell^w:\mathfrak g\to V$ is simply given by $X\mapsto X\cdot w$. Now let $\delta^r\phi$ be the right logarithmic derivative of the curve $\phi$, so $\delta^r\phi(t)=T_{\phi(t)}\rho^{\phi(t)^{-1}}(\phi'(t))$. Equivalently, this means that $\phi'(t)=T_e\rho^{\phi(t)}(\delta^r\phi(t))$. (In the notation you use, this would probably just be $\phi'=\delta^r\phi\cdot\phi$, but I find this rather tricky to use.)
Using this you can now compute
$$
T_{\phi(t)}\ell^{v(t)}(\phi'(t))=T_{\phi(t)}\ell^{v(t)}(T_e\rho^{\phi(t)}(\delta^r\phi(t)))
$$
From above, we know that $\ell^{v(t)}\circ \rho^{\phi(t)}=\ell^{\phi(t)\cdot v(t)}$ and differentiating this we get $T_{\phi(t)}\ell^{v(t)}\circ T_e\rho^{\phi(t)}=T_e\ell^{\phi(t)\cdot v(t)}$. Hence we can write the expression above as
$$
T_e\ell^{\phi(t)\cdot v(t)}(\delta^r\phi(t))=\delta^r\phi(t)\cdot \phi(t)\cdot v(t), 
$$
where the first dot denotes the infinitesimal action of $\mathfrak g$ on $V$ while the second denotes the action of $G$ on $V$. So the final result simply reads as
$$
\frac{d}{dt}(\phi(t)\cdot v(t))=\delta^r\phi(t)\cdot\phi(t)\cdot v(t)+\phi(t)\cdot v'(t). 
$$
