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There are two variables $o_1$ and $o_2$.

$$ o_1 = w_{11}x_1 + w_{12}x_2 \\ o_2 = w_{21}x_1 + w_{22}x_2 $$

Then is it true that $\frac{\partial ({o_1+o_2})}{\partial o_1} = 1$?

However I'm a bit confused about this.

If that true, let's say $y=2x$ and $z=x+y$, does that mean $\frac{\partial z}{\partial x} = 1$ is also true?

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Whenever you use partial differentiation, the coordinate system must be specified at the outset. Changes of variables must be handled by the multivariable chain rule.

Start by writing everything down as functions. If you have $f(\sigma_1,\sigma_2) = \sigma_1+\sigma_2$ then it's certainly the case that $\partial_{\sigma_1}f = 1.$ How you express this in terms of $x_1$ and $x_2$ is to write $$g(x_1,x_2)= f(\sigma_1(x_1,x_2),\sigma_2(x_1,x_2))$$ so that, say, $$ \frac{\partial}{\partial x_1}g(x_1,x_2) = \frac{\partial \sigma_1}{\partial{x_1}}\frac{\partial f}{\partial{\sigma_1}}+\frac{\partial \sigma_2}{\partial{x_1}}\frac{\partial f}{\partial{\sigma_2}}$$ where $\partial_{\sigma_1}f$ and $\partial_{\sigma_2}f$ are evaluated at $(\sigma_1(x_1,x_2),\sigma_2(x_1,x_2)).$

So in the case where $f(\sigma_1,\sigma_2) = \sigma_1+\sigma_2,$ the partial derivatives are both one and you should have $$ \frac{\partial}{\partial x_1}g(x_1,x_2) =\frac{\partial \sigma_1}{\partial{x_1}}+\frac{\partial \sigma_2}{\partial{x_1}}.$$

In your example, you say $z = x+y,$ which in my / your original notation corresponds to $f(\sigma_1,\sigma_2) = \sigma_1 +\sigma_2$. And then you say $y= 2x,$ which means we're taking $\sigma_1(x,y) = x$ and $\sigma_2(x,y) = 2x.$ (note this is not a change of variables, but the above all still holds.) So the formula gives $\partial_x g(x,y) = 3,$ not 1.

(Note this could be come to the answer more easily by substituting $x$ and $y$ in at the outset and noting $g(x,y) = 3x,$ but the chain rule works in more general cases)

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    It might be helpful to clarify that (if I'm understanding your example correctly) $\sigma_1=x$ and $\sigma_2= y(x)$. It's not exactly clear how your general answer (Which I think is quite good) lines up with the specific example of the OP. – 2017-01-07
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    Yes, I stuck with the notation OP used at the beginning, not the notation they used for their specific example. Thanks for the suggestion, will edit. – 2017-01-07
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There are two cases :

  • $w_{11}w_{22}-w_{12}w_{21}\neq 0$

$o_1$ and $o_2$ are independent, so

$\frac{\partial o_2}{\partial o_1}=0$

  • $w_{11}w_{22}-w_{12}w_{21}=0$

$o_2=\lambda o_1$ so $\frac{\partial o_2}{\partial o_1}=\lambda$.

for the second, we have $z=x+2x$ and $\frac{\partial z}{\partial x}=3$.