We have a function which takes a two-dimensional input and has two parameters. How to find (∂/∂(_2)) Question:
We have a function which takes a two-dimensional input $x = (x_1, x_2)$ and has two parameters
$w = (w_1, w_2)$ given by $f(x, w) = σ(σ(x1w1)w_2 + x_2)$ where $σ(x) = 1/(1+e^{-x}))$ . We use backpropagation to estimate the right parameter values. We start by setting both the parameters to
0. Assume that we are given a training point x1 = 1, x2 = 0, y = 5. Given this information
answer the next two questions. What is the value of $∂f/∂w_2$?
Solution:
Write $σ(x_1w_1)w_2 + x_2$ as $o_2$ and $x_1w_1$ as $o_1$
$∂f/∂w_2=∂f/∂o_2*∂o_2/∂w_2$
$∂f/∂w_2= σ(o_2)(1 − σ(o_2)) × σ(o_1)$ # Need to understand here
$∂f/∂w_2
= 0.5 ∗ 0.5 ∗ 0.5
=0.125$
Can some one help me to understand the solution?
What is the $f$ equation, which partially derivated with $o_2$ to get $σ(o_2)(1 − σ(o_2))$?
And not understood, from where $0.5$ came.
Please help. 
 A: The sigmoid function $\sigma(x)=[1+\exp(-x)]^{-1}$ has the following derivative:
$$ \frac{\partial \sigma}{\partial x} = \sigma(x)[1-\sigma(x)] \tag{1} $$
Let us now define 
\begin{align}
g(x,w) &= \sigma(x_1w_1)w_2 + x_2 \\
f(x,w) &= \sigma(g(x,w)) = \sigma( \sigma(x_1w_1)w_2 + x_2 )
\end{align}
Notice that $g$ is linear in $w_2$, so that:
$$ \frac{\partial g}{\partial w_2} = \sigma(x_1 w_1) \tag{2} $$
Using the chain rule, we get
\begin{align} 
\frac{\partial f}{\partial w_2} 
&= \frac{\partial \sigma}{\partial g} \frac{\partial g}{\partial w_2} \\[3mm]
&= \underbrace{\sigma(g(x,w))[1 - \sigma(g(x,w))]}_{\text{From} (1)} \;\underbrace{\sigma(x_1 w_1)}_{\text{From} (2)} \\
&= \sigma(o_2)[1 - \sigma(o_2)] \sigma(o_1)
\end{align}
where the last step uses $g(x,w)=: o_2$ and $x_1w_1=: o_1$.
The only confusing part is probably the derivative of the sigmoid function (which I linked to above). My favourite proof is this one by Hans Lundmark.

As for where the $0.5$ comes from, since $w_1=w_2=0$, $x_1=1$, and $x_2=0$, we have 
\begin{align}
o_1 &= x_1 w_1 = 0 \\
\sigma(o_1) &= \sigma(0) = [1 + \exp(0)]^{-1} = 2^{-1} = 0.5 \\
o_2 &= g(x,w) = \sigma(0)0 + 0 = 0\\
\sigma(o_2) &= \sigma(0) = 0.5 \\[2mm]
\therefore\;\;\; \frac{\partial f}{\partial w_2} &=  \sigma(0)[1 - \sigma(0)] \sigma(0) = 0.5[1-0.5]0.5= 0.5^3
\end{align}
