Honestly, I can't help you with your application, but I can tell you how the simple calculus works (which I'm assuming is what Alvaro is referring to).

To give an extremely simple introduction, a derivative is the instantaneous rate of change at a specific point on a line. The derivative of a function f(x) (which is notated f'(x)) is found by using the difference quotient:

(This picture's from the Wikipedia article on derivatives; just replace a with x.)

So if f(x) = x^{2}, f'(x) = {the limit as h approaches 0 of} ([x+h]^{2} - x^{2})/h. This may seem like it would equal 0; however, if the h variable is left in during expansion, you get:

{the limit as h approaches 0 of}(x^{2} + 2xh + h^{2} - x^{2})/h --> {the limit as h approaches 0 of}(2xh + h^{2})/h --> {the limit as h approaches 0 of} 2x + h = **2x.**

Therefore, the derivative of x^{2} is equal to 2x.

There are a lot of rules for derivatives that make derivation a lot easier, e.g. the power rule. This states that the derivative of a polynomial function's derivative can be found by taking each term, multiplying it by its power, and then reducing the power by 1. Thus the derivative of x^{3} = 3x^{2}, and the derivative of 2x^{2} + 3x = 4x + 3, etc.

Now, for Alvaro's function, (1+e^{-x})^{-1}, there are a few other rules involved; its derivative (I'm pretty sure) is -(1+e^{-x})^{-2}-e^{-x}.

Hope this helps,

Selenaut