My textbook Analysis II by Amann defines derivative as follows:
Let $E=(E,\|\cdot\|)$ and $F=(F,\|\cdot\|)$ are Banach spaces over the field $\mathbb{K}$; $X$ is an open subset of $E$. A function $f: X \rightarrow F$ is differentiable at $x_{0} \in X$ if there is an $A_{x_{0}} \in \mathcal{L}(E, F)$ such that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)-A_{x_{0}}\left(x-x_{0}\right)}{\left\|x-x_{0}\right\|}=0. $$ Then we write $\partial f(x_0)$ for $A_{x_0}$.
Could you please confirm if my below understanding is correct?
Assume that $f,g:X \to F$ are differentiable at $x_0 \in X$. Then there are $\partial f(x_0), \partial g(x_0) \in \mathcal{L}(E, F)$ such that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)-\partial f(x_0)\left(x-x_{0}\right)}{\left\|x-x_{0}\right\|}=0 $$ and $$ \lim _{x \rightarrow x_{0}} \frac{g(x)-g\left(x_{0}\right)-\partial g(x_0)\left(x-x_{0}\right)}{\left\|x-x_{0}\right\|}=0. $$
It follows that $$ \lim _{x \rightarrow x_{0}} \frac{(f+g)(x)-(f+g)\left(x_{0}\right)-(\partial f(x_0) + \partial g(x_0))\left(x-x_{0}\right)}{\left\|x-x_{0}\right\|}=0. $$
Because the sum of $2$ continuous linear maps is again continuous linear. Then $\partial (f+g)(x_0) = \partial f(x_0) + \partial g(x_0)$. As such, $$\partial (f+g)(x_0)(a+b) = \partial f(x_0)(a) + \partial g(x_0)(a) + \partial f(x_0)(b) + \partial g(x_0)(b)$$ for all $a,b \in X$.