# Can't understand the proof of the first backpropagation equation in Nielsen's neural network book

I am reading Proof of the four fundamental equations chapter of "Neural Networks and Deep Learning" and I think I got the general idea about backpropagation and the math involved, but there is this part which I can't figure out:

Let's begin with Equation (BP1), which gives an expression for the output error, $$\delta^L$$. To prove this equation, recall that by definition $$\begin{eqnarray} \delta^L_j = \frac{\partial C}{\partial z^L_j}. \tag{36}\end{eqnarray}$$ Applying the chain rule, we can re-express the partial derivative above in terms of partial derivatives with respect to the output activations, $$\begin{eqnarray} \delta^L_j = \sum_k \frac{\partial C}{\partial a^L_k} \frac{\partial a^L_k}{\partial z^L_j}, \tag{37}\end{eqnarray}$$ where the sum is over all neurons $$k$$ in the output layer. Of course, the output activation $$a^L_k$$ of the $$k^{\rm th}$$ neuron depends only on the weighted input $$z^L_j$$ for the $$j^{\rm th}$$ neuron when $$k=j$$. And so $$\partial a^L_k / \partial z^L_j$$ vanishes when $$k≠j$$. As a result we can simplify the previous equation to $$\begin{eqnarray} \delta^L_j = \frac{\partial C}{\partial a^L_j} \frac{\partial a^L_j}{\partial z^L_j}. \tag{38}\end{eqnarray}$$

I don't understand why, in the second equation, he is summing over all neurons $$k$$ in the output layer. Intuitively I'd have obtained the third equation directly from the first one by applying the chain rule, considering only the output activation $$a^L_j$$. I understand that $$C$$ is a function depending on all the output activations, so maybe the reason is because in its expression $$C = \frac{1}{2} \sum_j (y_j-a^L_j)^2$$ there is the summation. However, I feel like I am missing something.

That's because the author is considering in that part that the activation function of a neuron $$j$$ only depends on the value of $$z_j^L$$. This happens for example with the sigmoid or ReLU activation functions.
However, when using activation functions like Softmax (described in Chapter 3.1.4), we have that: $$a_j^L = \frac{e^{z_j^L}}{\sum_k e^{z_k^L}}$$ Here we can see that $$\partial a_j^L/\partial z_k^L \neq0$$ (or equivalently: $$\partial a_k^L/\partial z_j^L \neq0$$) for $$k\neq j$$, so the correct computation of $$\delta_j^L$$ is the equation given by the author: $$\delta_j^L=\sum_k\frac{\partial C}{\partial a_k^L}\frac{\partial a_k^L}{\partial z_j^L}$$
So, to sum up, the above equation is the general expresion of $$\delta_j^L$$, however if we use certain activation functions, it can be simplified to the equation $$(38)$$ of the question.
Yes, in this case, it's obvious. The author is simply writing a general equation for taking derivatives in a multivariate setting. If $$f(x,y)$$ is a function of $$x,y$$ and we look for the derivative of $$f$$ wrt $$t$$, it's customary to write the following: $$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$