Computing the Hessian of maximum log likelihood function

I am trying to find the Hessian matrix for the maximum log likelihood function given training data {(xi, yi)} for i=1:N with yi ∈ {+1, −1} for each i = 1, . . . , N for the function: When I try to find the Hessian matrix for this function, I get this: which gives a vector, not a number which doesn't seem be correct. How do you calculate the off diagonals of the Hessian matrix without the result being a vector?

This is what I get for the diagonals of the Hessian and the gradients: • as far as I can see you have only differentiated once with respect to the parameter (w,b) hence you get a vector. Then you need to differentiate again with respect to (w,b). Also there are other errors as I can see from a quick glance (might be wrong) but have another look. – Jesper Hybel Dec 5 '18 at 22:58
• The factors of $\log(10)$ definitely don't belong: "log" in the log likelihood formula means the natural log. – whuber Dec 5 '18 at 23:10

If you differentiate a multivariable function $$f$$ with respect to a vector $$\mathbf{w}$$ then you get a derivative that is itself a vector (the gradient vector). If you would like to get scalar second derivatives then you need to differentiate with respect to the elements of $$\mathbf{w}$$ instead of the whole vector. In this particular case, if you have a vector $$\mathbf{w} = (w_1,...,w_m)$$ then the Hessian matrix will consist of the following scalar second-order partial derivatives:
$$\mathbf{H}(b,\mathbf{w}) = \begin{bmatrix} \frac{\partial^2 f}{\partial b^2}(b, \mathbf{w}) & \frac{\partial^2 f}{\partial b \partial w_1}(b, \mathbf{w}) & \cdots & \frac{\partial^2 f}{\partial b \partial w_m}(b, \mathbf{w}) \\ \frac{\partial^2 f}{\partial b \partial w_1}(b, \mathbf{w}) & \frac{\partial^2 f}{\partial w_1^2}(b, \mathbf{w}) & \cdots & \frac{\partial^2 f}{\partial w_1 \partial w_m}(b, \mathbf{w}) \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial^2 f}{\partial b \partial w_m}(b, \mathbf{w}) & \frac{\partial^2 f}{\partial w_1 \partial w_m}(b, \mathbf{w}) & \cdots & \frac{\partial^2 f}{\partial w_m^2}(b, \mathbf{w}) \\ \end{bmatrix}.$$