Derivative of the logistic function
Show \sigma'(\eta)=\sigma(\eta)\bigl(1-\sigma(\eta)\bigr) for \sigma(\eta)=1/(1+e^{-\eta}), and that \sigma(-\eta)=1-\sigma(\eta). Use the first fact to confirm the Hessian -X^\top W X.
\sigma'(\eta) =\frac{e^{-\eta}}{(1+e^{-\eta})^2} =\frac{1}{1+e^{-\eta}}\cdot\frac{e^{-\eta}}{1+e^{-\eta}} =\sigma(\eta)\bigl(1-\sigma(\eta)\bigr),
since 1-\sigma(\eta)=e^{-\eta}/(1+e^{-\eta}). Also
\sigma(-\eta)=\frac{1}{1+e^{\eta}}=\frac{e^{-\eta}}{e^{-\eta}+1}=1-\sigma(\eta).
The score has ith contribution (y_i-p_i)x_i, with p_i=\sigma(\eta_i) and \eta_i=\beta^\top x_i. Differentiating again, \partial p_i/\partial\beta=p_i(1-p_i)x_i, so the Hessian is
-\sum_{i=1}^n p_i(1-p_i)\,x_i x_i^\top=-X^\top W X.
Concavity follows from that negative semi-definite Hessian. Existence of a finite optimum needs separate conditions.