'Algebraic' computation of the differential of a Lie algebra homomorphism (using vector fields) The following is an exercise from section 1.5.4 of Goodman and Wallach's textbook.
Let $$ \phi(A) = \begin{bmatrix} \det(A)^{-1} & 0 \\ 0 & A \end{bmatrix}$$ for $A \in GL(n,\mathbb{C})$.
Show that $$d\phi(X) =  \begin{bmatrix} -tr(X) & 0 \\ 0 & X \end{bmatrix}$$ for $X \in \mathfrak{g}$.
Using the definition of the differential given by the exponential map, $$ d \phi(X) = \frac{d}{dt} \phi ( \exp(tX)) \vert_{t=0},$$ then a straightforward application of $\det(\exp(X)) = \exp(\text{tr}(X))$ gives the desired identity.
On the other hand, the 'algebraic' definition of $d\phi$ using vector fields is the following.
For $A \in \mathfrak{gl}(V)$, define the derivation $X_A$ by $$X_A f (u) = \frac{d}{dt} f(u(I+tA))\vert_{t=0}$$ for $u \in G$ and $f \in \mathcal{O}[G]$.
The map $A \mapsto X_A$ gives an isomorphism between the Lie algebra of $G$ and the Lie algebra of left-invariant vector fields on $G$.
Then for any $A \in \mathfrak{g}$ and any regular function $f$ on $G$, we have $$ X_A (f \circ \phi) = (X_{d\phi(A)}f) \circ \phi.$$
I'm not sure though how to use this definition to compute $d\phi(X)$ as above.
 A: Let $f \in \mathcal{O}[\operatorname{SL}(n+1, \mathbb{C})], u \in \operatorname{GL}(n, \mathbb{C}), A \in \mathfrak{g}$ and $X_A$ the corresponding derivation. For convenience, let us denote $d\phi(X_A)$ by  $Y_A$, $\phi(u) = w$, and $h = f\circ \phi$. By definition, we have that
\begin{align}
Y_Af(w) :=&\ X_Ah(u)= \frac{d}{dt}h(u\cdot(I+tA))\bigg|_{t=0} \\
=&\ \frac{d}{dt}f\left( w\cdot \phi(I+tA)\right)\bigg|_{t=0}
\end{align}
since $\phi:\operatorname{GL}(n, \mathbb{C})\rightarrow \operatorname{SL}(n+1, \mathbb{C})$ is a group homomorphism. Recall that $f$ is a function of $(n+1)^2+1$ variables, i.e.
\begin{align}
f(\det(x)^{-1},x_{01},\ldots,x_{n0} ,x_{11}, \ldots, x_{n\ n}).
\end{align}
Then, by the chain rule of differential calculus, we have that
\begin{align}
\frac{d}{dt}f\left( \phi(u)\phi(I+tA)\right)\bigg|_{t=0} =&\ \frac{d}{dt}f\left( \begin{bmatrix}
\det(u)^{-1} & 0\\
0 & u
\end{bmatrix}
\begin{bmatrix}
\det(I+tA)^{-1} & 0\\
0 & I+tA
\end{bmatrix}
\right)\bigg|_{t=0}\\
=&\ \left(-\frac{\operatorname{tr}A}{\det(u)}\frac{\partial }{\partial \det(x)^{-1}}
+\sum_{i=1}^{n}\sum_{j=1}^n (u\cdot A)_{ij}\frac{\partial }{\partial x_{ij}}\right)f(w).
\end{align}
Furthermore, we can identify $Y_A = d\phi_u(X_A)$ with a $(n+1)\times (n+1)$ matrix just like how we identify $X_A$ with $A$, that is,
\begin{align}
d\phi_u(X_A) = 
\begin{bmatrix}
-\det(u)^{-1}\operatorname{tr}  A & 0\\
0 & u\cdot A
\end{bmatrix}.
\end{align}
Since we only care about $u=e$, then we have the desired result.
Edit: Notice that
\begin{align}
 f(\phi(u)\cdot \phi(I+tA))= f(x_{00},x_{01},\ldots,x_{n0} ,x_{11}, \ldots, x_{n\ n})
\end{align}
where
\begin{align}
x_{00} =&\ \det(u\cdot (I+tA))^{-1}, \ \ x_{0i} = x_{i0} =0 \text{ for }  i=1, \ldots, n,\\
x_{ij} =&\ (u\cdot (I+tA))_{ij}  \ \text{ for }\ i, j=1, \ldots, n.
\end{align}
Then, by chain rule, we have that
\begin{align}
\frac{d}{dt}f\left( \phi(u)\phi(I+tA)\right) =&\ \sum^n_{i=0}\sum^n_{j=0}\frac{\partial f}{\partial x_{ij}}\frac{dx_{ij}}{dt}
= \frac{\partial f}{\partial x_{00}}\frac{dx_{00}}{dt}+\sum^n_{i=1}\sum^n_{j=1}\frac{\partial f}{\partial x_{ij}}\frac{dx_{ij}}{dt}. 
\end{align}
Hence, it suffices to compute $d x_{ij}/dt$ and $d x_{00}/dt$. By Jacobi's formula, we see that
\begin{align}
\frac{d}{dt} \det(u\cdot (I+tA))^{-1} =&\ -\det(u\cdot (I+tA))^{-2}\frac{d}{dt} \det(u\cdot (I+tA))\\
=&\ -\det(u\cdot (I+tA))^{-2}\det(u) \frac{d}{dt}\det(I+tA)\\
=&\ -\det(u\cdot (I+tA))^{-2}\det(u)\operatorname{tr}\left(\operatorname{adj}(I+tA)A \right).
\end{align}
Finally, set $t=0$ yields
\begin{align}
\frac{d}{dt} \det(u\cdot (I+tA))^{-1}\bigg|_{t=0} = -\det(u)^{-1}\operatorname{tr}\left(\operatorname{adj}(I)A \right) =-\det(u)^{-1}\operatorname{tr}\left(A \right).
\end{align}
Next, notice that
\begin{align}
\frac{dx_{ij}}{dt} =&\ \frac{d}{dt} (u\cdot (I+tA))_{ij} = \frac{d}{dt} \sum^n_{k=1} u_{ik}(I+tA)_{kj}\\
=&\ \frac{d}{dt} \sum^n_{k=1} u_{ik}(\delta_{kj}+tA_{kj}) = \sum^n_{k=1} u_{ik}A_{kj} = (u\cdot A)_{ij}.
\end{align}
