Let $G$ be a Lie group and $L_a$ be the left multiplication map $x\mapsto ax$. Let $X\in \text{T}_e(G)$ be a fixed tangent vector, I wonder why the map
$$G \to \text{T}G, x \to d(L_x)_e(X)$$
is smooth.
I know I need to use the fact that the multiplication map is smooth somewhere. I have trouble in making use of the smooth structure of $\text{T}G$.
I´m gonna change your notation a little bit because is not very suggestive. Let´s represent the map as Vg := (dLg)(e)(X) where X is the vector at the tangent space to the identity, then: Vg := (dLg)(e)(X) = (dLg)(e)(r´(0)) = d(Lg∘r)(0) = d(μ(g;r(t)))(0) where r is a curve going trough the identity whose differential at the 0 is X and μ is the product map in the Lie group. So we know that the product map is smooth, so is particularly smooth in g, so also is it´s differential.