Q:

Differentiating this sum

This derivative comes up in the proof about the remainder of the Taylor expansion. 
This derivative is given and I can't follow the steps:

$$ \frac{d}{dy} R_{n+1}(y,x)=\frac{d}{dy}(f(x)-t_n(y,x))$$
$$ = - \sum\limits_{k=0}^n \frac{f^{k+1}(y)}{k!}(x-y)^k 
 + \sum\limits_{k=1}^n \frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1} 
 = - \frac{f^{n+1}(y)}{(n)!}(x-y)^{n} $$

So first, $f(x)$ is dropped, since we use $d/dy$ and $f(x)$ is a constant.
But how do we differentiate the n-th Taylor polynomial? 
I'm pretty certain the chain rule is involved and the fact that you can differentiate power series by using the power rule, but I can't seem to get the steps correct. 
Could somebody explain the steps used to differentiate  $\frac{d}{dy}(f(x)-t_n(y,x))$

A:

First:
$$\frac{d}{dy}f(x)=0$$
I suppose $t_n$ is the Taylor expansion until n:
$$t_n(x,y)=\sum\limits_{k=0}^n \frac{f^{(k)}(y)}{k!}(x-y)^k$$
The last is a sum of products, so:
$$\frac{d}{dy}t_n=\sum\limits_{k=0}^n \frac{d}{dy}\left( \frac{f^{(k)}(y)}{k!}\right)(x-y)^k+ \sum\limits_{k=0}^n \frac{f^{(k)}(y)}{k!}\frac{d}{dy}(x-y)^k$$
You can play with the $\sum$ and you are done.
ADDED:
(Corrected the superscript for the order of the derivative)
Differentiating the sum is to sum the derivatives, so is, differentiate each of:
$$\frac{1}{k!}·f^{(k)}(y)(x-y)^k$$
and sum. Each one but the first is the product of two functions of y, $f^{(k)}(y)$ and $(x-y)^k$ and a constant.
$$\frac{d}{dy}\left(\frac{1}{k!}·f^{(k)}(y)(x-y)^k\right)=\frac{1}{k!}\frac{d}{dy}\left(f^{(k)}(y)(x-y)^k\right)=$$
$$=\frac{1}{k!}\left(\frac{d}{dy}f^{(k)}(y)\right)(x-y)^k+\frac{1}{k!}f^{(k)}(y)\frac{d}{dy}(x-y)^k=$$
$$=\frac{1}{k!}f^{(k+1)}(y)(x-y)^k+\frac{1}{k!}f^{(k)}(y)(-1)k(x-y)^{k-1}=$$
$$=\frac{f^{(k+1)}(y)}{k!}(x-y)^k-\frac{k}{k(k-1)!}f^{(k)}(y)(x-y)^{k-1}=$$
$$=\frac{f^{(k+1)}(y)}{k!}(x-y)^k-\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}$$
Now, we have to sum the terms for each k being careful with the first, it is not a product: it is $f^{(1)}(y)$
$$\frac{d}{dy}t_n(y,x)=f^{(1)}(y)+\sum\limits_{k=1}^n\left(\frac{f^{(k+1)}(y)}{k!}(x-y)^k-\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}\right)=$$
$$=f^{(1)}(y)+\sum\limits_{k=1}^n\frac{f^{(k+1)}(y)}{k!}(x-y)^k-\sum\limits_{k=1}^n\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}=$$
$$=\sum\limits_{k=0}^n\frac{f^{(k+1)}(y)}{k!}(x-y)^k-\sum\limits_{k=1}^n\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}=$$
$$=\sum\limits_{k=1}^{n+1}\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}-\sum\limits_{k=1}^n\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}=$$
$$=\frac{f^{(n+1)}(y)}{n!}(x-y)^n+\sum\limits_{k=1}^{n}\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}-\sum\limits_{k=1}^n\frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}=$$
$$=\frac{f^{(n+1)}(y)}{n!}(x-y)^n$$
Finally:
$$\frac{d}{dy}(f(x)-t_n(y,x))=-\frac{f^{(n+1)}(y)}{n!}(x-y)^n$$

