I am trying to get this formula's derivative with respect to $\lambda_i$ but I can't. I tested with multiple online calculators but they fail to do so too. Can anyone help me?
In each iteration from $i$ to $n$, we have a $\lambda_i$ in numerator and sum of all lambdas in denominator including $i$-th lambda itself:
$$ e_k = \frac 1 2 \left( \sum_{i=1}^n b_{ki} \frac{e^{\lambda_i}}{\sum_{j=1}^n e^{\lambda_j}} - d_k \right)^2. $$
You just have to proceed step by step. I will denote by $\partial_i$ the partial derivative with respect to $\lambda_i$.
Let me first consider the quantity $$ E := \sum_{j=1}^n e^{\lambda_j}. $$ You have $\partial_i E = e^{\lambda_i}$.
Then you are interested in $$ e_k = \frac 1 2 \left( \sum_{\ell=1}^n \frac{b_{k \ell} e^{\lambda_{\ell}}}{E} - d_k \right)^2 $$
Hence $$ \begin{split} \partial_i e_k & = \left( \frac{b_{ki} e^{\lambda_i}}{E} - \frac{\partial_i E}{E^2} \sum_{\ell=1}^n b_{k\ell}e^{\lambda_\ell} \right) \left( \sum_{\ell=1}^n \frac{b_{k \ell} e^{\lambda_{\ell}}}{E} - d_k \right) \\ & = \frac{e^{\lambda_i}}{E} \left( b_{ki} - \sum_{\ell=1}^n \frac{b_{k\ell}e^{\lambda_\ell}}{E} \right) \left( \sum_{\ell=1}^n \frac{b_{k \ell} e^{\lambda_{\ell}}}{E} - d_k \right). \end{split} $$