Derivative of generalized Taylor expansion of a function between Banach Spaces 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.
 A: 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:

We have two Banach spaces $E,F$, a fixed vector $y \in E$, a continuous symmetric $k$-linear map $\omega: E^k \to F$. Now, we consider the map $E \to F$ defined by
  \begin{align}
x \mapsto \omega(x-y, \dots x-y).
\end{align}
  We wish to calculate its derivative at a point $x$.

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.
