Differential of a rational representation of algebraic groups Let $(\pi,V)$ be a rational (regular) representation of $G \subset GL(n,\mathbb{C})$, $\pi^*$ the dual representation, and set $\rho = \pi \otimes \pi^*$.
Let $T : W \otimes W^* \to \text{End}(W)$ denote the canonical isomorphism for any vector space $W$.
I know that for $Y \in \text{End}(W)$, $T \circ (Y \otimes I) = Y \cdot T$, and supposedly $T \circ (I \otimes Y^t) = T Y$ (this last identity is written in the textbook - Goodman & Wallach - but I'm not convinced it's correct).
If we set $$ \sigma(g) = T \rho(g) T^{-1},$$ then $(\sigma, \text{End}(V))$ is another rational representation of $G$, and $ \sigma(g) (Y) = \pi(g) Y \pi(g)^{-1}$.
Now I know that $d\rho(A) = d\pi(A) \otimes I - I \otimes d\pi(A)^t$, where $A \in \mathfrak{g}$ the Lie algebra of $G$.
Somehow I am supposed to see that $$ d \sigma(A) (Y) = d \pi(A) Y - Y d\pi(A),$$ but I cannot figure it out.
I have tried expanding the left hand side using the previous two identities, but this only yields
$$ d\sigma(A)(Y) = d\pi(A)(Y) - T d\pi(A) T^{-1}(Y),$$ which is close-but-not-quite-there.
I don't think going back to first principles using the definition of the differential via derivative of the exponential and/or vector fields will work here, but I cannot think of what else to try.
 A: $\newcommand{\End}{\operatorname{End}}$
I'm going to remove $T$ from consideration by instead defining actions of $\End(V)$ on the left and right of both $\End(V)$ and $V \otimes V^*$, which commute with $T$. The endomorphism $Y \in \operatorname{End}(V)$ acts on the left of the vector space $\End(V)$ by postcomposition $Y \cdot_L X = YX$, and on the left of $V \otimes V^*$ by action on the first term: $Y \cdot_L (v \otimes f) = Yv \otimes f$. As for right actions, we have $X \cdot_R Y = XY$ (precomposition), and $(v \otimes f) \cdot_R Y = v \otimes (f \circ Y) = v \otimes Y^t f$.
Now we have a group homomorphism $\pi \colon G \to \End(V)$ defining a representation, which we use to define another representation $\sigma: G \to \End(\End(V))$ by
$$\sigma(g)(X) = \sigma(g) \cdot_L X \cdot_R \sigma(g^{-1}) = \sigma(g) X \sigma(g^{-1}).$$
This action was defined using the left and right actions: left action by $\sigma(g)$, and right action by $\sigma(g^{-1})$. We can do the same to get a representation $\rho: G \to \End(V \otimes V^*)$ by
$$ \rho(g)(v \otimes f) = \sigma(g) \cdot_L (v \otimes f) \cdot_R \sigma(g^{-1}) = \sigma(g) v \otimes \sigma(g^{-1})^t f.$$
Note that $\sigma(g^{-1})^t = \sigma^*(g)$ by definition. While it is not so clear how to write down $\sigma(g)$ as an endomorphism of $\End(V)$, it is clear to write down $\rho(g)$ as an endomorphism of $\End(V \otimes V^*)$, we simply have $\rho(g) = \pi(g) \otimes \pi^*(g)$.
Knowing the rules for duals and tensor products, one can mechanically differentiate $\rho$ to get
$$ d\rho(g) = d \pi(g) \otimes 1 - 1 \otimes d \pi(g)^t,$$
which one could write down in terms of actions:
$$ d \rho(g)(v \otimes f) = d \pi(g) \cdot_L (v \otimes f) - (v \otimes f) \cdot_R d\pi(g).$$
Now just transport this action description back to $\End(V)$:
$$ d \sigma(g)(X) = d \pi(g) \cdot_L X - X \cdot_R d\pi(g) = d \pi(g) X - X d \pi(g).$$
