First, define Darboux derivative. There is one Lie group $G$ and one manifold $M$. Let $\phi:M\rightarrow G$ be a smooth map. The Darboux derivative $\Delta(\phi):TM \rightarrow M\times \mathfrak g$ of $\phi$ is given by
$\Delta(\phi)(X_x)=T(R_{\phi(x)^{-1}})(T(\phi)(X_x))$.
Second, suppose $V$ is a vector space. $f:M\rightarrow V$ and $\phi:M\rightarrow Hom(V,V)$ are both smooth maps. $\phi$ takes values in $GL(V)$. It is easy to get
$X(\phi(f))=X(\phi)(f)+\phi(X(f)),\quad X\in\mathfrak X(M)$
Here, $\phi(f)|_x=\phi(x)(f(x))$.
OK. Here comes my problem. It is said that $X(\phi)(f)=-\Delta(\phi)(X)(\phi(f))$. I tried to prove this relation, but I cannot get the sign correct. I will post my derivation:
Let's start with $\Delta(\phi)(X)(\phi(f))$. Suppose a curve $\gamma(t)$ in $M$ passing through a point, say $p=\gamma(t)\in M$ and $X$ tangent to it. We can evaluate it at $p$ in the following way
$\Delta(\phi)(X)(\phi(f))=\frac{d}{dt}\Big|_{t=0} [\phi(\gamma(t))\phi(\gamma(0))^{-1}]\phi(f(\gamma(0)))$
Here, I treat $\phi(x)$ as a matrix and the term behind $\frac{d}{dt}\Big|_{t=0}$ is a matrix multiplication. Further more, I get
$=\frac{d}{dt}\Big|_{t=0}[\phi(\gamma(t))]\phi(\gamma(0))^{-1}\phi(\gamma(0))(f(\gamma(0))) =\frac{d}{dt}\Big|_{t=0}[\phi(\gamma(t))]f(\gamma(0))=\frac{d}{dt}\Big|_{t=0}[\phi(\gamma(t))f(\gamma(0))]=X(\phi)(f)|_{\gamma(0)}$.
So where does the minus sign come from?