Question about Lemma 5.2.1 McDuff/Salamon Symplectic Topology I have a question about a formula in the proof of said lemma. 

Suppose we have a weakly Hamiltonian action of a connected Lie group $G$ on a symplectic manifold. Let $g \in G$. Since $G$ is connected, we find a path $\gamma \colon [0,1] \to G$ such that $\gamma(0) = e$ and $\gamma(1) = g$. Define $\eta \colon [0,1] \to \mathfrak{g}$ by $\eta(t) := \dot{\gamma}(t)\gamma(t)^{-1}$ (I suppose one should understand this as $\eta(t) = DR_{\gamma(t)^{-1}}(\dot{\gamma}(t))$ where $R_g \colon G \to G$ denotes multiplication by $g$ on the right). Then we have that $$\frac{d}{dt}\operatorname{Ad}_{\gamma(t)^{-1}}(\xi) = \operatorname{Ad}_{\gamma(t)^{-1}}[\xi,\eta(t)], \qquad \forall \xi \in \mathfrak{g}, t \in [0,1],$$ where $$\operatorname{Ad}_{g^{-1}}(\xi) := \frac{d}{dt}\bigg\vert_{t = 0}g^{-1}\exp(-t\xi)g$$ denotes the Adjoint action of $G$ on its Lie algebra $\mathfrak{g}$. Why does the above formula hold? 

My attempt: Since the adjoint action is a Lie algebra homomorphism, the rhs is equal to $$[\operatorname{Ad}_{\gamma(t)^{-1}}(\xi),\operatorname{Ad}_{\gamma(t)^{-1}}(\eta(t))],$$ but honestly I am stucked. Has anyone a hint for me how to show this?
 A: Evaluating the derivative given above, we find:
\begin{eqnarray*}
\frac{d}{dt} Ad_{\gamma^{-1}(t)}(\xi) & = & \frac{d}{dt} \left . \frac{d}{ds} \right |_{s=0} \gamma^{-1}(t)\exp(-\xi s) \gamma(t) \\
& = & \left .\frac{d}{ds}\right |_{s=0} \frac{d}{dt} \gamma^{-1}(t)\exp(-\xi s)\gamma(t) \\
& = & \left .\frac{d}{ds}\right |_{s=0} \left [\dot{\gamma}^{-1}(t)\exp(-\xi s) \gamma(t) + \gamma^{-1}(t)\exp(-\xi s) \dot{\gamma}(t)    \right ] \\
& = & \left .\frac{d}{ds}\right |_{s=0} \left [-\gamma^{-1}(t)\dot{\gamma}(t) \gamma^{-1}(t) \exp(-\xi s)\gamma(t) + \gamma^{-1}(t)\exp(-\xi s) \dot{\gamma}(t)    \right ] \\
& = & \left .\frac{d}{ds}\right |_{s=0} \left [-\gamma^{-1}(t)\dot{\gamma}(t) \gamma^{-1}(t) \exp(-\xi s)\gamma(t) + \gamma^{-1}(t)\exp(-\xi s) \dot{\gamma}(t)\gamma^{-1}(t)\gamma(t)    \right ] \\
& = & \left .\frac{d}{ds}\right |_{s=0} \gamma^{-1}(t)\left [\exp(-\xi s) \eta(t) - \eta(t) \exp(-\xi s)   \right ]  \gamma(t) \\ 
& = & \gamma^{-1}(t) \left[ -\xi \eta(t) + \eta(t)\xi   \right ] \gamma(t) \\
& = & \gamma^{-1}(t) (-[\xi, \eta(t)]) \gamma(t) \\
& = & \left . \frac{d}{ds} \right |_{s=0} \gamma^{-1}(t) \exp\left (-[\xi, \eta(t)]s\right ) \gamma(t) \\
& = & Ad_{\gamma^{-1}(t)} [\xi, \eta(t)].
\end{eqnarray*}
Here I used the identity that $\dot{\gamma}^{-1}(t) = -\gamma^{-1}(t)\dot{\gamma}(t)\gamma^{-1}(t)$ in the fourth line.
