How does an activation function's derivative measure error rate in a neural network? A blog post called "Text Classification using Neural Networks" states that the derivative of the output of a sigmoid function is used to measure error rates.

What is the rationale for this?

I thought the derivative of a sigmoid function output is just the slope of the sigmoid line at a specific point. 

Meaning it's steepest when sigmoid output is 0.5 (occuring when the sigmoid function input is 0).

Why does a sigmoid function input of 0 imply error (if i understand correctly)?

Source: https://machinelearnings.co/text-classification-using-neural-networks-f5cd7b8765c6


  We use a sigmoid function to normalize values and its derivative to
  measure the error rate. Iterating and adjusting until our error rate
  is acceptably low.


def sigmoid(x):
    output = 1/(1+np.exp(-x))
    return output

def sigmoid_output_to_derivative(output):
    return output*(1-output)

def train(...)
    ...
    layer_2_error = y - layer_2
    layer_2_delta = layer_2_error * sigmoid_output_to_derivative(layer_2)
    ...


UPDATE

Apologies. I don't think I was clear (I've updated the title)

I understand we don't need to use sigmoid as the activation funtion (we could use relu, tanh or softmax).

My question is about using the derivative to measure the error rate (full quotation from article above in yellow) -> what does the derivative of the activation function have to do with measuring/fixing the "error rate"?
This derivative is used when calculating the error of your machine learning algorithm during gradient based minimization methods. 
Read below for more info. 

When performing supervised classification (with X, Y data vectors of inputs and outcome data to train with) you begin with the error function 

E(X, Y; &theta;)= &sum;i (&fnof;(xi; &theta;)-yi)2

for total error over all data instances i, where f is your neural network, linear regression,...method of interest and &theta; is the set of weights. The goal here is to find weights that minimize your error when predicting training data (y) (which ideally generalizes to new data as well). To be explicit, &fnof;(xi; &theta;); outputs value of interest which should be yi. And E measures how far off it is in prediction. 

So to train your classifier, you optimize E with something like gradient descent. Thus when &part;E/&part;&theta; = 0 (for a particular &theta;), that means you hit a local minimum for the error function, or a point where the error in the current state of the predictor is low, meaning it is (hopefully) a good predictor.

Note the &fnof; here is not the same as an activation function, as a neural network is defined differently than in linear regression, etc. and must perform a special kind of gradient descent called backpropagation. 

So when you take &part;E/&part;&theta;, what does it equal for a neural net? You should note the activation functions derivative is involved which is how it’s used to measure error so to say.