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Assume an expression $$ E = \sum\limits_{i}\sum\limits_{j} \log a_{ij} - \sum\limits_{i=1}^j r_i\left(\sum\limits_{j} a_{ij} - 1\right) $$ I need to find following derivative $$ \frac{\partial E}{\partial a_{ij}} $$ So $$ \frac{\partial E}{\partial a_{ij}} = \frac{\partial \sum_{i}\sum_{j} \log a_{ij}}{\partial a_{ij}} - \frac{\partial \sum_{i=1}^j r_i (\sum_{j} a_{ij} - 1)}{\partial a_{ij}} \\ $$ Then $$ \frac{\partial \sum_{i}\sum_{j} \log a_{ij}}{\partial a_{ij}} = \frac{1}{a_{ij}}\\ $$ But I'm not really sure what to do with the second component. It feels like this expression in brackets should be equal to $1$ after differentiation, but I don't know how to prove that. It doesn't also help to write a few terms for i.e. $i = 2$ and $j = 3$, as there is $j$ in the upper limit in the first sum, and it confuses me a bit.

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It is not a problem of differentiation. Your expression $\sum_{i=1}^j r_i\left(\sum_{j} a_{ij} - 1\right)$ is not well-formed, for the reason you mention yourself.

If you replace it by $\sum_ir_i\left(\sum_{j} a_{ij} - 1\right)$ then its partial derivative with respect to $a_{i,j}$ will be $r_i.$

But if you insist on restricting your double sum to $i\le j,$ rather replace it by $\sum_ir_i\left(\sum_{j\ge i} a_{ij} - 1\right)$: then, its partial derivative with respect to $a_{i,j}$ will be $r_i$ if $i\le j,$ and $0$ if $i>j.$

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