Schaums's outline on Tensor Calculus has the following for problem 1.25: Calculate $\frac{\partial}{\partial x_\ell}(a_{ijk} x_i x_j x_k)$ given that the $a_{ijk}$ are constants.
Is the following a correct representation of the partial derivative in Summation notation?
$$\delta^i_\ell a_{ijk} x_j x_k + \delta^j_\ell a_{ijk} x_i x_k + \delta^k_\ell a_{ijk} x_i x_j$$
I believe that the $a_{ijk}$ cannot be pulled out because of their dependence on the different Kronecker delta components. If any of this is wrong, please explain.
The correct way to calculate $\frac{\partial}{\partial x_\ell}(a_{ijk} x_i x_j x_k)$ given that $a_{ijk}$ are constants is:
$$\frac{\partial}{\partial x_\ell}(a_{ijk} x_i x_j x_k) = a_{ijk} \frac{\partial}{\partial x_\ell}(x_i x_j x_k)$$
$$= a_{ijk} (\frac{\partial x_i}{\partial x_\ell} x_j x_k + \frac{\partial x_j}{\partial x_\ell} x_i x_k + \frac{\partial x_k}{\partial x_\ell} x_i x_j)$$
Given that $\frac{\partial x_p}{\partial x_q} = \delta_{pq}$,
$$= a_{ijk} (\delta_{i\ell} x_j x_k + \delta_{j\ell} x_i x_k +\delta_{k\ell} x_i x_j)$$
Redistributing leads to
$$= a_{ijk} \delta_{i\ell} x_j x_k + a_{ijk} \delta_{j\ell} x_i x_k + a_{ijk} \delta_{k\ell} x_i x_j$$
$$= a_{\ell jk} x_j x_k + a_{i \ell k} x_i x_k + a_{ij\ell} x_i x_j$$
Rewriting indices (both free and dummy) for independent terms leads to
$$= a_{ij\ell} x_i x_j + a_{i\ell j} x_i x_j + a_{k\ell ij} x_i x_j$$
$$= (a_{ij\ell} + a_{i\ell j} + a_{\ell ij}) x_i x_j$$