Simplifying partial derivative of cross-entropy function

Question

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: $$\begin{eqnarray} \frac{\partial C}{\partial w_j} = \frac{1}{n} \sum_x x_j(2\sigma(z)y - \sigma(z) - y). \tag{61}\end{eqnarray} $$

Answer 1

$$\begin{eqnarray} \frac{\partial C}{\partial w_j} = -\frac{1}{n} \sum_x \left( \frac{y(1-\sigma(z))}{\sigma(z)(1-\sigma(z))} -\frac{\sigma(z)(1-y)}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = -\frac{1}{n} \sum_x \left( \frac{y(1-\sigma(z))-\sigma(z)(1-y)}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = -\frac{1}{n} \sum_x \left( \frac{(y-y \sigma(z))-(\sigma(z)-y \sigma(z))}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = -\frac{1}{n} \sum_x \left( \frac{(y-y \sigma(z))+(-\sigma(z)+y \sigma(z))}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{-(y-y \sigma(z))+-(-\sigma(z)+y \sigma(z))}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{(-y+y \sigma(z))+(\sigma(z)-y \sigma(z))}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{-y+y \sigma(z)+\sigma(z)-y \sigma(z)}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{-y+y \sigma(z)-y \sigma(z)+\sigma(z)}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{-y+\sigma(z)}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{\sigma(z)-y}{\sigma(z)(1-\sigma(z))} \right)\sigma'(z) x_j. \\ = \frac{1}{n} \sum_x \left( \frac{(\sigma(z)-y)\sigma'(z) x_j}{\sigma(z)(1-\sigma(z))} \right). \\ = \frac{1}{n} \sum_x \left( \frac{(\sigma(z)-y)\sigma'(z) x_j}{\sigma(z)(1-\sigma(z))} \right). \\ = \frac{1}{n} \sum_x \left( (\sigma(z)-y) x_j \right). \\ = \frac{1}{n} \sum_x \left ( x_j(\sigma(z)-y) \right). \\ \end{eqnarray}$$

since $$ \sigma(z)(1-\sigma(z)) = \sigma'(z)$$