In order to find the parameters \( \theta_i \) we will then minimize the spread of \( \chi^2(\boldsymbol{\theta}) \) by requiring

$$ \frac{\partial \chi^2(\boldsymbol{\theta})}{\partial \theta_j} = \frac{\partial }{\partial \theta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\theta_0x_{i,0}-\theta_1x_{i,1}-\theta_2x_{i,2}-\dots-\theta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, $$

which results in

$$ \frac{\partial \chi^2(\boldsymbol{\theta})}{\partial \theta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\theta_0x_{i,0}-\theta_1x_{i,1}-\theta_2x_{i,2}-\dots-\theta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, $$

or in a matrix-vector form as

$$ \frac{\partial \chi^2(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\theta}\right). $$

where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \boldsymbol{b} \) with elements \( b_i = y_i/\sigma_i \).