Let $E$ and $F$ be Banach spaces and let $f: E \to F$ be a $n+1$ times differentiable function. We define for a given $y\in E$ the Taylor expansion of $f$ as the following: $$ T_n(x,y)=\displaystyle\sum_{k=0}^{n}{\dfrac{d^kf_y(x-y)^k}{k!}} $$ where in this case we use the abusive notation of $(x-y)^k$ as the $k$-tuple with all entries being $x-y$. We have that $T_n(x,y): E \to F$ with respect to $x$ (with $y$ and $n$ being constant).
Question. What is the derivative of this series at the point $y$? ($d_yT_n(x,y)=?$)
Here is what I've worked out so far. By linearity of the derivative the problem becomes one of finding the derivative of $d^kf_y(x-y)^k$ with respect to $x$, if I am not mistaken. In this case we have that $d^kf_y(x-y)^k$ is a multilinear map which means it has derivative equal to the map that sends a point $(s_1,s_2,....,s_k)$ to : $\displaystyle\sum_{i=1}^{k}{d^kf_y(x-y,...,s_i,...,x-y)}$. using the symmetry of higher derivatives we get that this derivative is equal to the map $k \cdot d^kf_y(x-y,....,x-y,s)$. I believe this leads to a form similar to the derivative of the Taylor polynomial for real functions. Is this the correct way to approach this problem? Am I taking the derivative on the wrong variable? Any help is appreciated.
Edit: I initially made a very silly mistake which led to me an incorrect answer. But you're exactly right.
Since taking derivatives is a linear operator, we can focus on simply taking the derivative with respect to $x$ of $x \mapsto d^kf_y(x-y)^k$. So, let's just forget about all the clutter and rewrite the problem with modified notation as follows:
To aid in this calculation, it might be clearer to introduce the following maps: $\sigma: E \to E$ defined by \begin{align} \sigma(x) := x-y, \end{align} the subtraction map, and also the "diagonal map" $\delta: E \to E^k$ \begin{align} \delta(x) &:= (x, \dots, x) \in E^k. \end{align} Then, the map we want to consider is $\omega \circ \delta \circ \sigma$. Let's calculate the derivative at a point $x$: \begin{align} d(\omega \circ \delta \circ \sigma)_x &= d \omega_{(\delta \circ \sigma)(x)} \circ d \delta_{\sigma(x)} \circ d \sigma_x \\ &= d \omega_{(\delta \circ \sigma)(x)} \circ \delta \circ \text{id}_E \tag{$\ddot{\smile}$} \end{align} where the last line is because $\delta$ and $\sigma$ are continuous linear and affine transformations respectively. We now recall that since $\omega$ is continuous and multilinear, \begin{align} d \omega_{(x_1, \dots, x_k)}(s_1, \dots, s_k) &= \sum_{i=1}^k \omega(x_1, \dots, \underbrace{s_i}_{i^{th} \text{ spot}}, \dots, x_k ) \end{align} So, now, let's evaluate $(\ddot{\smile})$ on a vector $s \in E$: \begin{align} (d \omega_{(\delta \circ \sigma)(x)} \circ \delta)(s) &= d \omega_{(x-y, \dots, x-y)}\left(s, \dots, s \right) \\ &= \sum_{i=1}^k \omega(x-y, \dots, \underbrace{s}_{i^{th} \text{ spot}}, \dots, x-y) \\ &= \sum_{i=1}^k \omega(x-y, \dots, x-y, s) \\ &= k \cdot \omega(x-y, \dots, x-y, s), \end{align} where in the second last line I used the symmetry of $\omega$.
So, finally lets put this all together for your original question. We have: \begin{align} d(T_{n}\left( \cdot, y)\right)_x(s) &= \sum_{k=0}^n \dfrac{k \cdot (d^kf)_y(x-y, \dots, x-y, s)}{k!} \end{align} The $k=0$ term is clearly $0$; I leave it to you to clean up the expression in whichever way you find fitting.