What is the gradient descent rule using binary cross entropy (BCE) with tanh? Similar to this post, I need the gradient descent step of tanh but now with binary cross entropy (BCE). 

So we have 

$$
\Delta \omega = -\eta \frac{\delta E}{\delta \omega}
$$

Now we have BCE:

$$
E = −(ylog(\hat{y})+(1−y)log(1−\hat{y}))
$$

Considering my output is $\hat{y} = tanh(\omega .x)$, $x$ is my input vector and $y_i$ is the corresponding label here. 
$$
\frac{\delta E}{\delta \omega} = \frac{\delta −(ylog(tanh(wx))+(1−y)log(1−tanh(wx)))}{\delta \omega} 
$$

Now on this website they do something similar for the normal sigmoid and arrive at (eq 60): 

$$
\frac{σ′(z)x}{ σ(z)(1−σ(z))}(σ(z)−y)
$$

Could we use that and continue there? We can get the derivative like this and get:

$$
\frac{tanh′(wx)x}{tanh(wx)(1−tanh(wx))}(tanh(wx)−y)
\\= \frac{x-xtanh(wx)^2}{tanh(wx)(1−tanh(wx))}(tanh(wx)−y)
\\= \frac{x-x\hat{y}^2}{\hat{y}(1−\hat{y})}(\hat{y}−y)
\\= \frac{(\hat{y} + 1)x(\hat{y} - y)}{\hat{y}}
$$

Wherever I look, I don't find this :)

Update

Given the first answer that gives $(1 + \hat{y})(1 - \hat{y})$, we arrive at the same

$$
\frac{tanh′(wx)x}{tanh(wx)(1−tanh(wx))}(tanh(wx)−y)
\\= \frac{x(1 + \hat{y})(1 - \hat{y})}{\hat{y}(1−\hat{y})}(\hat{y}−y)
\\= \frac{(\hat{y} + 1)x(\hat{y} - y)}{\hat{y}}
$$
Let 'a' be the output from an activation function like sigmoid or tanh.

Therefore, the derivative of sigmoid is a*(1-a) whereas the derivative for tanh is (1+a)*(1-a).

Just follow the derivation of sigmoid except replace the derivative of sigmoid with that of tanh.