$$\prod_i \frac{\prod_jexp(a_{ij}\theta- \delta_{ij})}{\sum_hexp(a_{ih}\theta- \delta_{ih})}$$
Taking log
$$\sum_i\sum_j {(a_{ij}\theta- \delta_{ij})}-{\sum_ilog\sum_hexp(a_{ih}\theta- \delta_{ih})}$$
Derivative with respect to $\theta$
$$\sum_i\sum_j {(a_{ij})}-{\sum_i\frac{(a_{ih}exp((a_{ih}\theta- \delta_{ih}))}{\sum_hexp(a_{ih}\theta- \delta_{ih})}}$$
The short answer is no, because taking $$ \log\left(\sum_h \mathrm{e}^{a_{ih}}\right) $$ can not be separated so I would expect an exponential in there e.g. in general $$ \log(a + b) \neq \log(a) + \log(b) $$
The solution would be $$ \sum_i\sum_j (\alpha_{ij}\theta -\delta_{ij}) - \sum_i\log\sum_h\left[\mathrm{e}^{\alpha_{ih}\theta -\delta_{ih}}\right] $$
