understanding batch normalization
Let's assume that you are trying to minimize the following loss for a given task, $$\ell(y, \hat{y}) = \frac{1}{2}\| y - \hat{y}\|^2, $$ where $\|
\cdot\|$ is the Eucledian distance, and the predicted output is $\hat{y} =\hat{x}$ for simplicity. The gradients then can be computed as follow: 

$$\Delta b = - \frac{\partial\ell}{\partial\hat{x}} \cdot \frac{\partial\hat{x}}{\partial b}, \hspace{20pt} \Delta \omega = - \frac{\partial\ell}{\partial\hat{x}} \cdot \frac{\partial\hat{x}}{\partial \omega}$$

Now, the gradient of $\hat{x}$ with respect to the bias $b$ is $$\frac{\partial\hat{x}}{\partial b} = \frac{\partial}{\partial b}(x-E[x]) = \frac{\partial}{\partial b}\left((u+b)-E[(u+b)]\right) = 1 - \frac{\partial}{\partial b}E[(u+b)]$$ 

Ignoring the fact $E(x)$ is dependant on $b$ makes the above gradient equals to 1, and hence keeps updating the bias as follow: 

$$\frac{\partial\hat{x}}{\partial b} = 1 -  \frac{\partial}{\partial b}E[x] = 1 -0 
 = 1$$ 
and then 

$$\Delta b = - \frac{\partial\ell}{\partial\hat{x}} \cdot (1), \hspace{20pt} b \leftarrow b + \Delta b $$

Otherwise, if you wanna consider this dependency, the gradient becomes 0, and hence no update as follow: 

$$\frac{\partial\hat{x}}{\partial b} = 1 -  \frac{\partial}{\partial b}E[u+b] =  1 -  \left(\frac{\partial}{\partial b}E[u]+\frac{\partial}{\partial b}E[b]\right) 
 = 1 - (0+1) = 0$$ 
and then 
 $$\Delta b = - \frac{\partial\ell}{\partial\hat{x}} \cdot (0), \hspace{20pt} b \leftarrow b + 0 $$ 

In both cases, regardless bias update, the loss function will remain fixed,  

$$u+(b+\Delta b)−E[u+(b+∆b)]=u+b−E[u+b], $$

however, in the first case, the bias will grow indefinitely.