Yeah but not for any function, given that f(x)=y, f(2x)=2y. Which is what backpropagation tries to do (find x such that f(x) is something). Anyhow we're talkin

Yeah but not for any function, given that f(x)=y, f(2x)=2y. Which is what backpropagation tries to do (find x such that f(x) is something). Anyhow we're talking past each other. I mean that it's linear in a very strict/specific sense. Take a fixed newtork with a fixed loss function, for a particular set of inputs and particular target outputs (think 1 batch, or even just one example). Now as you backprop you continuously compute derivatives of what's currently on the layer with respect to the loss value, starting from the loss function, then the bottom most layer, then the previous layer and so on. Now this can be decomposed into a series of matrix multiplication (chain rule), where yes, in any particular run the matrices depend on the weights, inputs and outputs in a nonlinear fashion. But the overall effect is just a composition of linear operators (matrix multiplication). That's why things go to shit when the network is deep and the derivative matrices are all contracting or all expanding. And just to connect with your example: if f=d(error)/dW then yes d(2error)/dW=2f.