# What is derivative in sigmoid function about?

sigmoid > sigmoid(x): 1/(1+np.exp(-x))
derivate > deriv(x): sigmoid(x)*(1-sigmoid(x))

x = 10

sigmoid = 0.999954602131
derivate = 0.196616057953


I think I am correct about calculations. But can someone please explain to me, like to a five year old what derivate value tells us at all? I just know it is a change of something, but change of what? What does 0.196616057953 has to do with 0.999954602131?

What confuses me even more... If i put any number from -x to x, derivate is always about 0.19 up to 0.25 What is the point of this derivate?

• Something may be wrong with your implementation: $0.999954602131 \times (1-0.999954602131) \approx 0.0000453958$ so when $x=10$ the derivative is far from $0.19$. The derivative is between $0.19$ and $0.25$ roughly when $-1.0718 \le x \le 1.0718$ but not otherwise – Henry Mar 1 '18 at 17:20
• It tells you the rate of change. Geometrically, what is the slope of the tangent line (numerically, it's a secant line unless you have the exact form of the derivative) to the sigmoid at a point? – Adrian Keister Mar 1 '18 at 18:19

The derivative is to the function what speed is to distance.

To make it simpler, for $x=10$ you got $sigmoid = 0.999954602131$.

Repeat the calculation for $x=9.9$ to get $sigmoid = something$.

Now compute $$\frac{ 0.999954602131-something}{10 -9.9}=10(0.999954602131-something)$$

This should look like the number you got for $derivate$

Edit

After Henry's comment, what it seems is that you computed $\text{sigmoid}(10)=0.9999546021$ but $\text{deriv}(1)=0.1966119332$ while $\text{deriv}(10)=0.0000453958$.