# Understanding the algebra behind a specific partial derivative equation

I am following this article about neural networks.

Given:

Until here I understand everything, but then he continues to:

I don't understand how he got to that conclusion. I think he skipped some algebra steps that would have made it easier for me to understand.

## 1 Answer

We know that:

(1) $$\frac{\partial}{\partial x}\big (f(x) + g(x) \big) = \frac{\partial}{\partial x}f(x) + \frac{\partial}{\partial x}g(x)$$

(2) $$\frac{\partial}{\partial x}a = 0$$

Now,

\begin{align*} &\frac{\partial}{\partial w_{12}^{1}} (w_{11}^{1}h_1^{2} + w_{12}^{1}h_2^{2} + w_{13}^{1}h_3^{2} + b_1^{1}) = & \text{[using (1)]}\\ &\frac{\partial}{\partial w_{12}^{1}} (w_{11}^{1}h_1^{2}) + \frac{\partial}{\partial w_{12}^{1}} (w_{12}^{1}h_2^{2}) + \frac{\partial}{\partial w_{12}^{1}} (w_{13}^{1}h_3^{2}) + \frac{\partial}{\partial w_{12}^{1}} b_1^1) = & \\\\ & \text{since } h_1^2 \text{ is independent of } w_{12}^1 \text{ , } h_3^2 \text{ is independent of } w_{12}^1 \text{ , } b_1^1 \text{ is independent of } w_{12}^1 & \text{[using (2)]}\\ & = 0 + \frac{\partial}{\partial w_{12}^{1}} (w_{12}^{1}h_2^{2}) + 0 + 0 \implies \frac{\partial}{\partial w_{12}^{1}} (w_{12}^{1}h_2^{2}) \end{align*}