Going from the differential to the derivative (Frechet and matrix calculus)

For a function $$f: A \rightarrow B$$, Frechet differentiability tells us that we want to find a linear operator that satisfies

$$\lim_{H\rightarrow 0} \frac{||f[X+H] - f[X] - G[H]||}{||H||} = 0$$

This would mean that $$G$$ is a good approximation of the change in $$f$$ at $$X$$ for some small $$H\in A$$. That is for the operator $$df: A \rightarrow B$$

$$df(X)[H] = G[H]$$

Wikipedia says that $$G$$ is defined as the Frechet derivative of $$f$$ at $$X$$. But I have a litte trouble connecting this to the traditional notion of the derivative, where we have a fraction i.e. something like $$\frac{dy}{dx}$$.

For example, consider a standard formula from the matrix cookbook

$$\frac{dTr(XA)}{dX} = A^T$$

The Frechet differentiability definition gets me up to

$$dTr(XA)[H] = G[H] = Tr(HA).$$

What is then done is \begin{align} dTr(XA)[H] &= Tr(HA) \\ &= A^T :H, \end{align}

where we just use the notation $$A:B = Tr(A^TB)$$. What is the correct way to go from here to the conclusion that $$\frac{dTr(XA)}{dX} = A^T$$

and what does the LHS even represent exactly since it's not a fraction in the traditional sense?

More generally, what is involved in going from the differential form i.e. $$df(X)[H] = G[H]$$ to the derivative form $$G = \frac{df(X)}{dX}$$?

If $$f: A\to B$$ is Frechet differentiable then for all $$X,V\in A$$ the directional derivative

$$d_Vf(X)=\lim_{t\to 0}\frac{f(X+tV)-f(X)}{t}$$

exists and $$d_Vf(X)=df(X)(V)$$.

Now if $$A=\mathbb R^{n\times n}$$, $$B=\mathbb R$$ and $$G = \frac{df(X)}{dX}$$ then $$G_{i,j}$$ is just the directional derivative $$d_{H_{i,j}}f(X)$$ where $$H_{i,j}$$ is the matrix which has a $$1$$ on the $$(i,j)$$-th position and is zero otherwise. So the relation is

$$G_{ij}=df(X)(H_{i,j})$$

$$\bullet$$ Now consider the special map $$f:\mathbb R^{n\times n}\to\mathbb R$$ $$X\mapsto Tr(XA)$$. As this map is already linear we have $$df(X)=f$$ for all $$X\in\mathbb R^{n\times n}$$ so aplying the above relation yields

$$G_{i,j}=df(X)(H_{i,j})=f(H_{i,j})=Tr(H_{i,j}A)=A_{j,i}$$

so $$G=A^T$$ which also can be computed directly.