For Lasso we need now, keeping a constraint on \( \vert\theta_0\vert+\vert\theta_1\vert=1 \), to take the derivative of the absolute values of \( \theta_0 \) and \( \theta_1 \). This gives us the following derivatives of the cost function
$$ C(\boldsymbol{\theta})=(4-2\theta_0)^2+(2-\theta_1)^2+\lambda(\vert\theta_0\vert+\vert\theta_1\vert), $$ $$ \frac{\partial C(\boldsymbol{\theta})}{\partial \theta_0}=-4(4-2\theta_0)+\lambda\mathrm{sgn}(\theta_0)=0, $$and
$$ \frac{\partial C(\boldsymbol{\theta})}{\partial \theta_1}=-2(2-\theta_1)+\lambda\mathrm{sgn}(\theta_1)=0. $$We have now four cases to solve besides the trivial cases \( \theta_0 \) and/or \( \theta_1 \) are zero, namely