I would like to find the derivative of this function with respect to $\alpha$.
$$K(\beta_r | \alpha) = \frac{e^{\alpha^{'}z(\beta_r)}}{\sum_{s \in S }e^{\alpha^{'}z(\beta_s)}}$$
By using the quotient rule, I have got:
$$\frac{\partial K(\beta_r|\alpha)}{\partial \alpha}=\frac{\left( \sum_{s \in S }e^{\alpha^{'}z(\beta_s)} \right)z(\beta_r) e^{\alpha^{'}z(\beta_r)}}{\left( \sum_{s \in S }e^{\alpha^{'}z(\beta_s)} \right)^2} - \frac{e^{\alpha^{'}z(\beta_r)} \left( \sum_{s \in S }z(\beta_s)e^{\alpha^{'}z(\beta_s)} \right)}{\left( \sum_{s \in S }e^{\alpha^{'}z(\beta_s)} \right)^2}$$
This doesn't seem to be right to me :-(.
You are correct, but the result can be simplified to $$ \frac{\partial K(\beta_r|\alpha)}{\partial \alpha} = \frac{e^{\alpha' z(\beta_r)}} {\left( \sum_{s \in S} e^{\alpha^{'}z(\beta_s)} \right)^2} \sum_{s \in S} \left[z(\beta_r) - z(\beta_s) \right] e^{\alpha'z(\beta_s)} $$