# implementing forward and backward of a Linear model I'm implementing the code of this abstraction.

The forward is easy and looks like that: I don't understand the backward path and how it fit's the abstraction in the first image: 1. Why is db defined as multiplication of ones of x's shape and dout ?
2. Why is dw defined as multiplication of ones of x.T and dout ?
3. Why both of them are accumulated. i.e it is used += and not = ?
4. Why is dw defined as multiplication of ones of dout and w.T ?

1. This is because the derivative wrt $$b$$ is $$1$$: $$\frac{\partial E}{\partial b} = 1$$
2. dout is the derivative of loss function wrt prediction. Using chain rule, $$\frac{dE}{dw} = \frac{dE}{dy}\frac{dy}{ds}\frac{ds}{dw}$$ The last term is the vector of input features $$x$$. In your case dout is the combination of the first two terms. For example, for MSE loss and sigmoid activation dout $$= (y-L)y(1-y)$$
4. For MLPs, you need to compute gradients for coarse layers using gradients of deep layers. For example, for MLP with one hidden layer with features $$\mathbf{z}$$ (hence 3 in total) vector of gradients wrt weights in the input layer $$\mathbf{w}^0$$ would be $$y= \sigma(\sum_kw^1_k \cdot\sigma(\sum_jw^0_jx_j))\\ \frac{\partial E}{\partial \mathbf{w^0}} = \frac{\partial L}{\partial y} \frac{\partial y}{\partial s} \frac{\partial s}{\partial \mathbf{z}}\frac{\partial \mathbf{z}}{\partial \mathbf{w}^0} = \frac{\partial E}{\partial \mathbf{z}}\frac{\partial \mathbf{z}}{\partial \mathbf{w}^{0}}\\ \frac{\partial E}{\partial \mathbf{w^0}} = (y-L) y(1-y) \sum_j\frac{\partial s}{\partial z_j}\frac{\partial z_j}{\partial \mathbf{w^0}} = (y-L) y(1-y) \sum_j\frac{\partial s}{\partial z_j}\frac{\partial z_j}{\partial s_j}\sum_i \frac{\partial s_j}{\partial w_{ij}}\\ \frac{\partial E}{\partial \mathbf{z}} = (y-L)y(1-y)\frac{\partial s}{\partial \mathbf{z}} = (y-L)y(1-y)\mathbf{w}^1$$ So, in other words, in order to compute gradients for weights in the input layer, you need gradients wrt neurons in the hidden layer