Neural Network Continuous Perceptron Error Function

Question

Let $u={\bf w}^{\text T}{\bf x}$ and $o=f(u)$. Can someone explain to me step by step how the partial derivative below was obtained? Thank you. $$\begin{eqnarray}\frac{\partial E({\bf w})}{\partial{\bf w}} & = & \frac12\frac{\partial(d-o)^2}{\partial{\bf w}}\\ & = & -(d-o)\frac{\partial f(u)}{\partial{\bf w}}\\ & = &-(d-o)\frac{\partial f(u)}{\partial u}\frac{\partial u}{\partial{\bf w}} \end{eqnarray}$$

Answer 1

I assume that $d$ is a constant with respect to $\bf{w}$.

For the second line, use the fact that $\frac{\partial}{\partial\bf{w}}(d-o)^2=2(d-o)\frac{\partial}{\partial\bf{w}}(d-o)$.

Then use $\frac{\partial}{\partial\bf{w}}(d-o)=\frac{\partial}{\partial\bf{w}}d-\frac{\partial}{\partial\bf{w}}o=0-\frac{\partial}{\partial\bf{w}}f(u)$.

For the third line, use the chain rule.