Consider problem $(P)$ with all functions continuously differentiable. Suppose that for each $b \in \mathbb{R}^m$, the optimal value $\phi(b)$ is attained and $\lambda^*$ is the corresponding Lagrange multiplier at the optimum. Then \begin{align*} \frac{\partial \phi}{\partial b_i} = \lambda_i^*. \end{align*}