Gradient computation, result verification I have a problem with the computation of the gradient of the function
$$
L(w) = -\dfrac{1}{N}\sum\limits_{n=1}^N y_{n}\log\left( \sigma(w^{T}x_{n}) \right) + (1 - y_{n})\log\left( 1-{\sigma}(w^{T}x_{n}) \right)
$$
where $\sigma$ is a sigmoid function defined by $\sigma(x) = \dfrac{1}{1 + \mathrm{e}^{-x}}$.
With my attempts I end up with the gradient taken with respect to $w$:
$$
\triangledown_{w}L(w) = -\dfrac{1}{N}\sum\limits_{n=1}^N x_{n}y_{n} -x_{n}\sigma(w^{T}x_{n}) \;.
$$
My process of computation was as follows: derivative of $\log\left( \sigma(w^{T}x_{n}) \right)$, then inner function of $\log$, so it's $\sigma$ and then inner function of $\sigma$ so $w^Tx$. The same, of course, for the second part of the sum.
Can you spot any obvious mistakes? Maybe I am not allowed to treat vectors ($w$ and $x$) like normal variables?
Thanks in advance.

An easy way to check is to check the components.
Recall that
$$
\frac{\partial}{\partial x}\sigma(x) = \sigma(x)[1-\sigma(x)]
\;\;\;\&\;\;\;
\frac{\partial}{\partial w_i} w^Tx_j = x_{ji}
$$
Then, the $i$th component of the gradient is:
\begin{align}
\frac{\partial}{\partial w_i} L(w)
&=
\frac{-1}{N}\sum_j y_j\frac{\sigma(w^Tx_j)[1-\sigma(w^Tx_j)]x_{ji}}{\sigma(w^Tx_j)}
+(1-y_j)\frac{(-1)\sigma(w^Tx_j)[1-\sigma(w^Tx_j)]x_{ji}}{1-\sigma(w^Tx_j)}\\
&=
\frac{-1}{N}\sum_j y_j[1-\sigma(w^Tx_j)]x_{ji} - (1-y_j)\sigma(w^Tx_j)x_{ji}\\
&=
\frac{-1}{N}\sum_j y_jx_{ji} - \sigma(w^Tx_j)x_{ji} 
\end{align}
Therefore, combining the components into one vector:
$$
\nabla L(w) = \frac{-1}{N}\sum_j x_j[y_j - \sigma(w^Tx_j)]
$$
as expected.
