Derive the derivative of cost function of logistic regression. I am trying to derive the derivative of the loss function of a logistic regression model.
Instead of 0 and 1, y can only hold the value of 1 or -1, so the loss function is a little bit different.   The following is how I did it. The answers I found online were all a little bit different from mine. I'd be grateful if anyone could help and see if I did something wrong!
\begin{align*}
            L &= -\sum_{n=1}^{N} \log \sigma\left( t_n (w^\top x_n + w_0) \right)\\
            \frac{dL}{dw} &=-\frac{d}{dw}\sum_{n=1}^{N} \log \sigma\left( t_n (w^\top x_n + w_0) \right)\\
            &=-\sum_{n=1}^{N} \frac{d}{dw}\log \sigma\left( t_n (w^\top x_n + w_0) \right)\\
            &=-\frac{d}{dw}\log \sigma\left( w^{'\top} X^{'\top} T \right)
        \end{align*}
where $w^{'} = 
    \begin{bmatrix} 
    w \\
    w_0
    \end{bmatrix}$, and
$x^{'}=
    \begin{bmatrix}
    -x_1^\top-, 1\\
    ...\\
    -x_n^\top-, 1\\
    \end{bmatrix}$
Now let $A(x)=log(x)$, $B(x)=\sigma(x)$, $C(x)= w^{'\top} X^{'\top}T$
Then,
\begin{align*}
        \frac{dL}{dw^{'}}&=\frac{dA(B)}{dB} \times \frac{dB(C)}{dC} \times \frac{dC}{dw^{'}}\\
        &=\frac{1}{B} \times \sigma(C)(1-\sigma(C)) \times \frac{dC}{dw^{'}}\\
        &=(1-\sigma(C)) \times X^{'\top}T\\
        &=(1-\sigma(w^{'\top} X^{'\top}T)) \times X^{'\top}T
    \end{align*}

$
\def\b{\omega_0}\def\s{\sigma}\def\o{{\tt1}}\def\p{\partial}
\def\L{{\cal L}}
\def\LR#1{\left(#1\right)}
\def\diag#1{\operatorname{diag}\LR{#1}}
\def\Diag#1{\operatorname{Diag}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
$Let $\o$ denote the all-ones vector and $T$ a diagonal matrix constructed from the $\{t_n\}$ values and define the vector
$$\eqalign{
p &= T\LR{Xw+\o\,\b} \qiq dp = TXdw \\
}$$
and the elementwise logistic function
$$\eqalign{
s &= \s(p) = \frac{e^p}{\o+e^p} \\
S &= \Diag{s} &\qiq ds=\LR{S-S^2}dp \\
}$$
Use the above notation to rewrite the loss function
and calculate its gradient.
$$\eqalign{
L &= -\o:\log(s) \\
dL &= -\o:S^{-1}ds \\
 &= -\o:S^{-1}\LR{S-S^2}dp \\
 &= \LR{S-I}\o:dp \\
 &= \LR{s-\o}:\LR{TXdw} \\
 &= X^TT\LR{s-\o}:dw \\
\grad{L}{w} &= X^TT\LR{s-\o} \\
}$$
This almost matches your result, except for the overall sign and the order of the factors.

In some of the steps above a colon is used to denote the matrix inner product
$$\eqalign{
A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\
A:A &= \|A\|^2_F \\
}$$
The properties of the underlying trace function allow the terms in such a product to be rearranged in many different but equivalent ways, e.g.
$$\eqalign{
A:B &= B:A \\
A:B &= A^T:B^T \\
C:\LR{AB} &= \LR{CB^T}:A \\&= \LR{A^TC}:B \\
}$$
Also note that my definition of the $X$ matrix omits the rightmost column of ${\large\tt1}s\:$ since $\,\b$ drops out of the derivative anyway.
