[Optimization]Doubt about Lagrangian multiplier's derivative

[zero064]

The Lagrangian dual equation is like the following:

$\displaystyle \displaystyle \psi(\lambda)\, =\, \min_x\, T(\mathbf{x})\, =\, f (\mathbf{x}) \, +\, \sum_{j\, =\, 1}^{\ell}\, \lambda_j \, h_j\, +\, \dfrac{1}{2}\, r\, \sum_{j\, =\, 1}^{\ell}\, \left[h_j(\mathbf{x})\right]^2$

This is T function.

$\displaystyle T(\mathbf{x})\, =\, f (\mathbf{x}) \, +\, \sum_{j\, =\, 1}^{\ell}\, \lambda_j \, h_j\, +\, \dfrac{1}{2}\, r\, \sum_{j\, =\, 1}^{\ell}\, \left[h_j(\mathbf{x})\right]^2$

Therefore, its total derivative with respect to lambda i should be:

$\displaystyle \dfrac{d \psi}{d \lambda_i}\, =\, \dfrac{\partial \psi}{\partial \lambda_i}\, +\, \nabla \psi^T\, \dfrac{d \mathbf{x}}{d \lambda_i}$

Because the gradient of T with respect to x is zero, the second term vanishes. But I don't know why the final result is this:

$\displaystyle \dfrac{d \psi}{d \lambda_i}\, =\,h_i (\mathbf{x}(\lambda)),\,$. . .$\displaystyle i\, =\,1,\, 2,\,...,\, \ell$

In my thought, the x is also function of lamda. It should depend on lamda. So the result of partial derivative should be

$\displaystyle h_i(\mathbf{x}(\lambda))\, +\, \lambda_i\, \dfrac{d}{d \lambda_i}\, (h_i)$

Why the second term vanishes? What's wrong with my idea? My classmate also asked the professor this question, but he didn't answer.