I have difficulty concluding the derivatives of the following:
$$y=\ln\bigg(\sum_{j=1}^ng_j\cdot e^{xE_j}\bigg)$$ and $$y=x\cdot \sum_{j=1}^n \bigg(g_j \cdot \frac{N}{\sum_{j=1}^ng_j\cdot e^{xE_j}}\cdot e^{xE_j}\cdot E_j\bigg)$$ Note that $x$ does not change during the summation. Are these possible to solve? If so, how?
EDIT: I have made a mistake when copying the second formula, I have corrected it.
If $y =\ln\bigg(\sum_{j=1}^ng_j\cdot e^{xE_j}\bigg) $ then, since $(\ln(f(x))' =\dfrac{f'(x)}{f(x)} $,
$\begin{array}\\ y' &=\dfrac{(\sum_{j=1}^ng_j e^{xE_j})'}{\sum_{j=1}^ng_je^{xE_j}}\\ &=\dfrac{\sum_{j=1}^ng_jE_j e^{xE_j}}{\sum_{j=1}^ng_je^{xE_j}}\\ \end{array} $
For your second $y$, you need to make the index in the inner sum different than the index in the outer sum.
$\begin{array}\\ y &=x \sum_{j=1}^n \bigg(g_j \dfrac{N}{\sum_{k=1}^ng_ke^{xE_k}} e^{xE_j}\bigg)\\ &=N\sum_{j=1}^n \bigg(g_j \dfrac{xe^{xE_j}}{\sum_{k=1}^ng_ke^{xE_k}} \bigg)\\ &=N\sum_{j=1}^n g_jy_j(x)\\ \end{array} $
where $y_j(x) = \dfrac{xe^{xE_j}}{\sum_{k=1}^ng_ke^{xE_k}} $.
Then
$\begin{array}\\ y_j'(x) &= \left(\dfrac{xe^{xE_j}}{\sum_{k=1}^ng_ke^{xE_k}}\right)'\\ &= \dfrac{(\sum_{k=1}^ng_ke^{xE_k})(xe^{xE_j})'-(\sum_{k=1}^ng_ke^{xE_k})'(e^{xE_j})}{(\sum_{k=1}^ng_ke^{xE_k})^2}\\ &= \dfrac{(\sum_{k=1}^ng_ke^{xE_k})((xE_j+1)e^{xE_j})-(\sum_{k=1}^ng_kE_ke^{xE_k})(e^{xE_j})}{(\sum_{k=1}^ng_ke^{xE_k})^2}\\ &= e^{xE_j}\dfrac{(xE_j+1)\sum_{k=1}^ng_ke^{xE_k}-\sum_{k=1}^ng_kE_ke^{xE_k}}{(\sum_{k=1}^ng_ke^{xE_k})^2}\\ \end{array} $
Now put this in $y'(x) =\sum_{j=1}^n g_jNy_j'(x) $, do any possible simplifications, correct any errors I may have made, and you are done.