Pontryagin's principle partial derivative for an electrical circuit

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Electrical circuit

What would be the partial derivative result for the following Hamiltonian used for Pontryagin's Minimum Principle? The Hamiltonian is:

$H = (U-u_{m})^2 +\lambda_1 \frac{di_L}{dt} $

The state-space equation for this system is: $\frac{di_L}{dt}=\frac{1}{L_m}(u_m - i_L R_m-u_l)$, $ U, R_m, L_m$ are constant values and $u_m$ is a variable voltage source.

And the partial derivative with respect to the state-space variable $i_L$ should be:

$\frac{\partial}{\partial i_L}(H)= 2(U-u_m)\frac{\partial u_m}{\partial i_L} + \lambda_1 \frac{\partial}{\partial i_L}\frac{di_L}{dt} = 2(U-u_m)\frac{\partial}{\partial i_L} (L_m\frac{di_L}{dt}+ i_L R_m+u_l)+\lambda_1 \frac{\partial}{\partial i_L}\frac{1}{L_m}(u_m - i_L R_m-u_l)$ $\frac{\partial}{\partial i_L}(H)= 2(U-u_m)R_m - \lambda_1 \frac{R_m}{L_m} $

But if I partially derive every component in the brackets, I get the following result:

$\frac{\partial}{\partial i_L}(u_m)=\frac{\partial}{\partial i_L}(L_m\frac{di_L}{dt})+\frac{\partial}{\partial i_L}(i_L R_m)+\frac{\partial}{\partial i_L}(u_L)=-R_m+R_m-R_m=-R_m$

I am kind of confused about how to interpret these results because, while every variable is a function of time, they are also mutually related with regards to basic el. circuit analysis. Intuitively, the equation $\frac{\partial}{\partial i_L}(u_m)=R_m$ should be correct because a change in current $i_L$ value should only change the value of $u_m$ by a factor $R_m$.