Let $V$ be the vector space $M_{n,n}(\bf{R})$, and suppose $g : V \to V$ is defined by $g(\mathbf{a}) = \mathbf{a}^{3}$. I'm asked to compute the derivatives of $g$ in the form $g'(\mathbf{a})(\mathbf{b})$ where $\mathbf{a}$ is in the domain of g and $\mathbf{b}$ is in the image.
My thought is to parametrize the matrix $\mathbf{a}$, $g(\mathbf{a}(t)) = \mathbf{a}(t)^3$ such that $\mathbf{a}(0) = \mathbf{a_0}$, and $\mathbf{a}'(0)=v$. Then, to find the derivative: $$Dg (\mathbf{a_0},v) =\frac{d}{dt}\bigg|_{t=0} g(\mathbf{a}(t))= 3\mathbf{a_0}^2 v $$
Is it right to infer from this statement that: $$ Dg(\mathbf{a_0})=3\mathbf{a_0}^2 : V \to \mathbf{R^n}, v\mapsto 3\mathbf{a_0}^2 v $$
Also, what's the best way to show that the function is differentiable? It seems pretty straightforward that if $\mathbf{a} \in V$ then the function is continuous, but is there a simple way to show that rigorously?
Given matrices $A$ and $B$ we have $$ \frac{(A+tB)^3-A^3}{t}=\frac{tA^2B+tABA+t^2AB^2+tBA^2+t^2BAB+t^2B^2A+t^3B^3}{t} $$ and so $$ \lim_{t\to0}\frac{(A+tB)^3-A^3}{t}=A^2B+ABA+BA^2. $$ In other words, $$ (D_Ag)B=A^2B+ABA+BA^2. $$