Context
In the derivation of the Boltzmann factor and the canonical partition function based essentially on Lagrange multipliers presented here, the equalities, \begin{align*} p_j &= \frac{1}{Z} e^{\frac{\lambda_2 E_j}{k_B}} \\ Z &= \sum_j e^{\frac{\lambda_2 E_j}{k_B}} \end{align*} are established, where $E_j$ refers to the energy of a microstate, $k_B$ is the Boltzmann constant, and $\lambda_2$ is the remaining undetermined multiplier. Then, using the Gibbs entropy $S = - k_B \sum_j p_j \ln(p_j)$ along with some algebra, \begin{gather*} S = -\lambda_2 U + k_B \ln(Z) \end{gather*} Finally, the article suggests the derivative, \begin{gather*} \frac{\partial S}{\partial U} = - \lambda_2 \end{gather*} without further work. Note that the article also improperly uses a total derivative instead of a partial here as the whole point is to make the identification $\frac{\partial S}{\partial U} = - \lambda_2 = \frac{1}{T}$.
My Question
My question comes from the details of working out the partial derivative. I obtained, \begin{align*} \frac{\partial S}{\partial U} &= - \lambda_2 - U \frac{\partial \lambda_2}{\partial U} + \frac{k_B}{Z} \frac{\partial Z}{\partial U} \\ &= - \lambda_2 - U \frac{\partial \lambda_2}{\partial U} + \frac{1}{Z} \sum_j e^{\frac{\lambda_2 E_j}{k_B}} \left[ E_j \frac{\partial \lambda_2}{\partial U} + \lambda_2 \frac{\partial E_j}{\partial U} \right] \\ &= - \lambda_2 + \frac{1}{Z} \sum_j e^{\frac{\lambda_2 E_j}{k_B}} \lambda_2 \frac{\partial E_j}{\partial U} \end{align*} where we have used the statistical definition of $U$ and the equalities above to cancel one of the derivative terms involving $\frac{\partial \lambda_2}{\partial U}$. It is at this point that I have a question. Clearly, for this to work, we must have $ \frac{\partial E_j}{\partial U} = 0$. But I cannot seem to really justify this step to myself. Is it simply that the energies of the particular microstates do not explicitly depend on the average energy? How would one justify this?