I will classify using a neural network algorithm. I use 2 output, Y1=1 (positive) and Y2=0 (negative). The architecture is as follows:
loss that I use is binary cross entropy with the following formula:
From the loss, the error gradient with respect to weight v is:
based on the architecture that I demonstrated, is it correct to calculate the error gradient with respect to v11 like this?
The substitution you made is correct. The formula you use is also correct assuming sigmoid activation at each layer:
According to the architecture, the forward and backward propagation equations are: $$\begin{align}Z_j&=\sigma\left(\sum_{i=1}^{n_{input}} X_iV_{ij}+b_j\right)\rightarrow \frac{\partial Z_j}{\partial V_{ij}}=X_iZ_j(1-Z_j) \\Y_k&=\sigma\left(\sum_{j=1}^{n_{hidden}} Z_jW_{jk}+b_k\right)\rightarrow \frac{\partial Y_k}{\partial Z_j}=Y_k(1-Y_k)W_{jk} \\\frac{\partial Y_k}{\partial V_{ij}}&=\sum_{j'=1}^{n_{hidden}}\frac{\partial Y_k}{\partial Z_{j'}}\frac{\partial Z_{j'}}{\partial V_{ij}}=\frac{\partial Y_k}{\partial Z_{j}}\frac{\partial Z_j}{\partial V_{ij}}=Y_k(1-Y_k)W_{jk}X_iZ_j(1-Z_j) \\E&=-\sum_{k=1}^{n_{out}}Q_k\log Y_k+(1-Q_{k})\log(1-Y_k)\rightarrow \frac{\partial E}{\partial Y_k}=\frac{Y_k-Q_k}{Y_k(1-Y_k)}\end{align}$$
Using the chain rule, we have $$\begin{align}\frac{\partial E}{\partial V_{ij}}&=\sum_{k=1}^{n_{out}}\frac{\partial E}{\partial Y_k}\frac{\partial Y_k}{\partial V_{ij}}=\sum_{k=1}^{n_{out}}(Y_k-Q_k)W_{jk}X_iZ_j(1-Z_j)\end{align}$$
And, substituting $i=j=1$ here yields the following: $$\begin{align}\frac{\partial E}{\partial V_{11}}&=\sum_{k=1}^2(Y_k-Q_k)W_{1k}X_1Z_1(1-Z_1)\\&=(Y_1-Q_1)W_{11}X_1Z_1(1-Z_1)+(Y_2-Q_2)W_{12}X_1Z_1(1-Z_1)\end{align}$$
