Derivative of left-multiplication in $GL(n,\mathbb{R})$ Given two maps between sets $f \colon X \to Y$ and $g \colon Y \to Z$, we define the pullback of $g$ by $f$ as $f^*g := g \circ f$.
Given a smooth map between manifolds $F \colon M \to N$ of dimensions $m$ and $n$, respectively, we define its differential (or derivative) at $p \in M$ by a map $d_pF \colon T_pM \to T_{F(p)}N$ given by $$d_pF(X_p) := X_pF^*,$$ that is, $d_pF(X_p)f = X_p(f \circ F)$ where $f \in C^\infty_p(M)$.
A curve in a manifold $M$ is a smooth map $c \colon I \to M$ where $I$ is an open interval in $\mathbb{R}$ containing $0$. The velocity of $c$ at $t_0 \in I$ is the tangent vector given by $$c'(t_0) := d_{t_0}c\left(\left.\frac{d}{dt}\right|_{t_0}\right) \in T_{c(t_0)}M.$$
Using these definitions, we can prove the following theorem:

Let $F \colon M \to N$ be a smooth map of manifolds, $p \in M$ and $X_p \in T_pM$. If $c$ is a curve starting at $p$ (that is, $c(0) = p$) in $M$ with velocity $X_p$ at $p$ (i.e., $c'(0) = X_p$), then $$d_pF(X_p) = \left.\frac{d}{dt}\right|_0 (F \circ c)^*(t).$$

Knowing about this result and about these definitions (that I wrote here so that we all are in the same terms, since there is a billion ways to define these concepts), I need help proving:

Prove that the derivative of the left multiplication by $g \in GL(n,\mathbb{R})$, $l_g \colon GL(n,\mathbb{R}) \to GL(n,\mathbb{R})$, is also left multiplication by $g$.

I'm stuck, i've tried writing the definition and do some calculations by I always get nowhere.
 A: The map $l_g$ is the restriction to $GL(n,\Bbb R)$ of the linear map$$\begin{array}{rccc}l_g\colon&\Bbb R^{n\times n}&\longrightarrow&\Bbb R^{n\times n}\\&M&\mapsto&gM.\end{array}$$And when you differentiate a linear map $f$ at any point of its domain, you always get $f$ once again.If you really want to apply that theorem, take $p\in GL(n,\Bbb R)$, take $v\in\Bbb R^{n\times n}$, and take a curve $c\colon(-\varepsilon,\varepsilon)\longrightarrow GL(n,\Bbb R)$ such that $c(0)=p$ and that $c'(0)=v$. Then$$(\forall t\in(-\varepsilon,\varepsilon)):l_g(c(t))=gc(t),$$and, since multiplication by $g$ is a linear map,$$\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}gc(t)=g.\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}c(t)=g.v.$$So, $D_pl_g(v)=g.v=l_g(v)$.
A: 
Then$$(\forall t\in(-\varepsilon,\varepsilon)):l_g(c(t))=gc(t),$$and, since multiplication by $g$ is a linear map,$$\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}gc(t)=g.\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}c(t)=g.v.$$So, $D_pl_g(v)=g.v=l_g(v)$.

To clarify this point. For any linear function $L$ the derivative as $t$ goes to $0$ is explicitly given by lim (L(c(t))-L(c(0))/t . Since $L$ is linear, we can rearrange it as L((c(t)-c(0))/t) for all t. And because we know linear functions are continuous the limit may be pulled in, giving L(c'(0)).
Another way to say this is (A(v+hw)-A(v))/h = (A(v)+A(hw)-A(v))/h=(hA(w))/h = A(w) and apply limit as t goes to 0 to both sides.
