Let $G$ be a real Lie group with identity $e$. Let $x_0\in G$ and $U\subset G$ be a submanifold containing $x_0$. Consider the map $$\varphi:G\times U\rightarrow G,\quad \varphi(g,u):=gug^{-1}.$$
We identify the Lie algebra $\mathfrak{g}$ of $G$ as the set of left invariant vector fields on $G$, so that isomorphic to the tangent space $T_eG$ at $e$. For each $g\in G$, denote by $l_g$ the left translation map $l_g:G\rightarrow G$, $l_g(x)=gx$; then for any $x\in G$, we identify $\mathfrak{g}$ with $T_xG$ via the isomorphism $dl_x|_e:T_eG\rightarrow T_xG$. Now for $x_0\in U$, we could identify $T_{x_0}U$ as a subspace of $T_{x_0}G$, thus a subspace $\mathfrak{u}$ of the Lie algebra $\mathfrak{g}$.
Now the question is: I want to compute the tangent map of $\varphi$ at $(g_0,x_0)$, namely $$d\varphi|_{(g_0,x_0)}:\mathfrak{g}\times \mathfrak{u}\rightarrow \mathfrak{g}.$$ I can guess that the correct answer should be $$d\varphi|_{(g_0,x_0)}: (X,Y)\mapsto {\rm Ad}(g_0)\left[ ({\rm Ad}(x_0)^{-1}-1)X+Y\right],$$ where ${\rm Ad}$ means the adjoint action of the Lie group on the corresponding Lie algebra. I think that I need this "$({\rm Ad}(x_0)^{-1}-1)$" to explain the "translation". But I don't know how to write down a precise proof. Can anyone help me to formalize a precise proof?