Showing a function is Frechet Differentiable？

I'll rewrite the definition of $H$ as $$ H(X) = GW F(X^T)^T + SX + C. $$ Let's assume that $F$ is Frechet differentiable at a particular point $X^T$, so that $$ F(X^T + \Delta X^T) = F(X^T) + F'(X^T) \Delta X^T + e(\Delta X), $$ and the error term $e(\Delta X)$ satisfies $$ \lim_{\Delta X \to 0} \frac{\|e(\Delta X)\|}{\| \Delta X \|} = 0. $$ Notice that \begin{align} H(X + \Delta X) &= GW F(X^T + \Delta X^T)^T + SX + S\Delta X + C \\ &= GW \left( F(X^T) + F'(X^T) \Delta X^T + e(\Delta X) \right)^T + SX + S \Delta X + C \\ &= \underbrace{GWF(X^T)^T + SX + C}_{H(X)} + \underbrace{GW(F'(X^T) \Delta X^T)^T + S \Delta X}_{H'(X) \Delta X} + \underbrace{GW e(\Delta X)^T}_{\text{small}}. \end{align}

Comparing this with the equation $$ H(X + \Delta X) \approx H(X) + H'(X) \Delta X $$ suggests that $H$ is differentiable at $X$ and that $H'(X)$ is the linear transformation defined by $$ \tag{1} H'(X) \Delta X = GW(F'(X^T) \Delta X^T)^T + S \Delta X. $$ To prove that this is true, we only need to show that $$ \tag{2} \lim_{\Delta X \to 0} \frac{\| GW e(\Delta X)^T \|}{ \| \Delta X \|} = 0 $$ To establish (2), let $L$ be the linear transformation defined by $$ L(v) = GW v^T. $$ Then \begin{align} \frac{\| GW e(\Delta X)^T \|}{ \| \Delta X \|} &= \frac{\| L(e(\Delta X)) \|}{\| \Delta X \|} \\ &\leq \frac{\| L \| \|e(\Delta X) \|}{\| \Delta X \|} \end{align} which approaches $0$ as $\Delta X \to 0$.

In order to reach the conclusion that $H$ is differentiable at $X$, we needed to assume that $F$ is differentiable at $X^T$.

I don't see a simpler way to express $H'(X)$, but maybe somebody else will.