Derivative wrt. to Lie bracket. Let $\mathbf{G}$ be a matrix Lie group, $\frak{g}$ the corresponding Lie algebra,
$\widehat{\mathbf{x}} = \sum_i^m x_i G_i$ the corresponding hat-operator ($G_i$ the $i$th basis vector of the tangent space/Lie algebra $\frak{g}$) and $(\cdot)^\vee$ the inverse of $\widehat{\cdot}$:
 $$(X)^\vee := \text{ that } \mathbf{x} \text{ such that } \widehat{\mathbf{x}} = X.$$
Let us define the Lie bracket over $m$-vectors as: 
$[\mathbf{a},\mathbf{b}] = \left(\widehat{\mathbf{a}}\cdot\widehat{\mathbf{b}}-\widehat{\mathbf{b}}\cdot\widehat{\mathbf{a}}\right)^\vee$.
(Example: For $\frak{so}(3)$, $[\mathbf{a},\mathbf{b}] = \mathbf{a}\times \mathbf{b}$ with $\mathbf{a},\mathbf{b} \in \mathbb{R}^3$.)
Is there a common name of the derivative:
$$\frac{\partial [\mathbf{a},\mathbf{b}]}{\partial \mathbf{a}}$$?

I may be misunderstanding your question, but it seems like you are asking for the derivative of the map
$$
\mathfrak g \to \mathfrak g, ~~ a \mapsto [a, b]
$$
where $b \in \mathfrak g$ is fixed.  Since $\mathfrak g$ is a vector space, the derivative at a point can be viewed as a map $\mathfrak g \to \mathfrak g$.  But the above map is linear so it is it's own derivative at any point.
