I am trying to derive a decision rule for a logistic regression function that has been localized with a Gaussian kernel. I know that I need to have the probability at one side $P(Y=1\mid X)$ or some notation along those lines but I am stuck on the final step.
\begin{align*} & \operatorname{min} \sum_{i=1}^{n} k \left( \frac{x-X_i}{h}\right) \left(y_i \beta_0(x) -\operatorname{log}(1+\operatorname{exp}(\beta_0(x))\right)) &\\ l(\beta_0(x)) & = \sum_{i=1}^{n} k \left( \frac{x-X_i}{h}\right) \left(y_i \beta_0(x) -\operatorname{log}(1+\operatorname{exp}(\beta_0(x))\right)) &\\ \frac{\partial l(\beta_0(x))}{\partial \beta_0(x)} & = \sum_{i=1}^{n} \quad [(0)](y_i\beta_0(x) - \operatorname{log}(1+\operatorname{exp}(\beta_0(x))) + k \left( \frac{x-X_i}{h}\right) \left[y_i - \frac{1}{1+\operatorname{exp}(\beta_0(x))} \times \operatorname{exp}(\beta_0(x))\right] &\\ 0 & = \sum_{i=1}^{n} k \left( \frac{x-X_i}{h}\right)\left[y_i - \frac{\operatorname{exp}(\beta_0(x))}{1+\operatorname{exp}(\beta_0(x))}\right] &\\ \text{where} &\\ & \frac{\operatorname{exp}(\beta_0(x))}{1+ \operatorname{exp}(\beta_0(x))} = p(x) &\\ \text{For class 1:} &\\ & \sum_{i=1}^{n} k \left( \frac{x-X_i}{h}\right)\left[y_i - p(x) \right] \geq 0 &\\ \text{For class 2:} &\\ & \sum_{i=1}^{n} k \left( \frac{x-X_i}{h}\right)\left[y_i - p(x)\right] < 0 &\\ \end{align*}
Summary
I am asking if someone could show me how to conclude the derivation above. I know the answer is binary but don't know much about derivations with kernels.