Solve $\frac{d}{da_k}\big[\sum_{k=1}^{p}{-a_k\cdot y(n-k)}\big]$

Question

$\frac{d}{da_k}\hat{y}(n) = \frac{d}{da_k}\big[\sum_{k=1}^{p}{-a_k\cdot y(n-k)}\big]$ is a formula that is part of this paper (p. 6, Formula (6)). It says the solution is: $\frac{d}{da_k}\hat{y}(n) = -y(n-k)$. BUT isn't the solution $\frac{d}{da_k}\hat{y}(n) = \sum_{k=1}^{p}{-y(n-k)}$ ?

Answer 1

I guess there is a little propblem concerning the differential operator. My guess is that the $a_k$ there refers to one of the indices, lets call it $k'$ with the restriction $1 \leq k' \leq p$, and so the sum becomes

$$\frac{d}{da_{k'}}\hat{y}(n)~=~\frac{d}{da_{k'}}\left[\sum_{k=1}^p -a_k\cdot y(n-k)\right]~=~\frac{d}{da_{k'}} \left[ -a_1\cdot y(n-1)-a_2\cdot y(n-2)-~\cdots - a_{k'}\cdot y(n-k')-\cdots -a_{p-1}\cdot y(n-{p-1})-a_p\cdot y(n-p)\right]$$

By differentiation of each of the sum elements with respect to $a_{k'}$ leads to

$$0 + 0 + \cdots -y(n-k') + \cdots + 0 + 0 $$

To put this all together yields to

$$\frac{d}{da_{k'}}\hat{y}(n)~=~-y(n-k')$$

The problem is that the indices of the sum and one single variable are denoted with the same letter.

Otherwise, if the differential would be part of every single sum element, your first guess would be right.