Derivative involving Summation Notation

[Lucid Dreamer]

Hello, I am looking at a derivation that involves (note x is a column vector)
\frac {d(\vec{x}^T\vec{x})} {d\vec{x}} = \vec{x}^{T}

So I convert to summation notation and evaluate as follows
\sum_{i,j} \frac {d(x_{i}x^{i})} {dx^{j}}
\sum_{i,j} \frac {dx_{i}} {dx^{j}} x^{i} + \sum_{i,j} x_{i} \frac {dx^{i}} {dx^{j}}
\sum_{i,j} \frac {dx_{i}} {dx^{j}} x^{i} + \sum_{i,j} x_{i} \delta_{ij}
\sum_{i,j} \frac {dx_{i}} {dx^{j}} x^{i} + \sum_{i} x_{i}

So I do recover the answer in the second term, but I am not sure what the derivative in the first term means. This derivative is of a row vector with respect to a column vector.

 

[mathman]

I am not familiar with the notation here, but I suspect dx_i/dx_j = 0 for i ≠ j.

 

[slider142]

I have only encountered this notation in metric spaces.
You will need to write the covector field F_i(x) = x_i as a function of the vector field Gⁱ(x) = xⁱ. This depends on the inner product of the vector space you are working in. If the induced metric of your space is M, then x_i = M_ijx^j. Thus, (d/dx^j)(x_i) = (d/dx^j)(M_ikx^k). If the components of the metric M_ik are constants, then we have (d/dx^j)(M_ikx^k) = M_ik(d/dx^j)(x^k) = M_ikδ^k_j = M_ij.
If your metric is the standard Euclidean metric, then M_ij = δ_ij, so
\sum_{i, j} \frac{dx_i}{dx^j}x^i = \sum_{i, j} \delta_{ij} x^i = \sum_j x_j
If your metric is not standard Euclidean, you replace it as above.