Using the matrix-vector expression for Lasso regression, we have the following cost function
$$ C(\boldsymbol{X},\boldsymbol{\theta})=\frac{1}{n}\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\theta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\theta})\right\}+\lambda\vert\vert\boldsymbol{\theta}\vert\vert_1, $$Taking the derivative with respect to \( \boldsymbol{\theta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicity)
$$ \frac{d \vert \theta\vert}{d \theta}=\mathrm{sgn}(\theta)=\left\{\begin{array}{cc} 1 & \theta > 0 \\-1 & \theta < 0, \end{array}\right. $$we have that the derivative of the cost function is
$$ \frac{\partial C(\boldsymbol{X},\boldsymbol{\theta})}{\partial \boldsymbol{\theta}}=-\frac{2}{n}\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\theta})+\lambda sgn(\boldsymbol{\theta})=0, $$and reordering we have
$$ \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\theta}+\lambda sgn(\boldsymbol{\theta})=\boldsymbol{X}^T\boldsymbol{y}. $$This equation does not lead to a nice analytical equation as in Ridge regression or ordinary least squares. We have absorbed the factor \( 2/n \) in a redefinition of the parameter \( \lambda \). We will solve this type of problems using libraries like scikit-learn and using our own gradient descent code in project 1.