How do I simplify:
$$\begin{eqnarray} \frac{\partial C}{\partial w_j} & = & -\frac{1}{n} \sum_x \left( \frac{y }{\sigma(z)} -\frac{(1-y)}{1-\sigma(z)} \right) \frac{\partial \sigma}{\partial w_j} \tag{58}\\ & = & -\frac{1}{n} \sum_x \left( \frac{y}{\sigma(z)} -\frac{(1-y)}{1-\sigma(z)} \right)\sigma'(z) x_j. \tag{59}\end{eqnarray}$$
to
$$\begin{eqnarray} \frac{\partial C}{\partial w_j} & = & \frac{1}{n} \sum_x \frac{\sigma'(z) x_j}{\sigma(z) (1-\sigma(z))} (\sigma(z)-y). \tag{60}\end{eqnarray}$$
to get:
$$\begin{eqnarray} \frac{\partial C}{\partial w_j} = \frac{1}{n} \sum_x x_j(\sigma(z)-y). \tag{61}\end{eqnarray} $$
where $$\sigma(z) = \frac{1}{1+e^{-zx}}$$
?
I end up with: $$\begin{eqnarray} \frac{\partial C}{\partial w_j} = \frac{1}{n} \sum_x x_j(2\sigma(z)y - \sigma(z) - y). \tag{61}\end{eqnarray} $$