y=(e^x_ Derivative of e ^ - x) / (e ^ x + e ^ - x)

y=(e^x_ Derivative of e ^ - x) / (e ^ x + e ^ - x)

Up and down city e ^ x
y=(e^2x-1)/(e^2x+1)
=(e^2x+1-2)/(e^2x+1)
=1-2/(e^2x+1)
So y '= - 2 * [- 1 / (e ^ 2x + 1) & # 178;] * (e ^ 2x + 1)'
=4e^2x/(e^2x+1)²

The derivative of y = E / x

y=e/x=ex^(-1),
y'=-ex^(-2)=-e/x^2

Finding the derivative of y = e ^ x
Detailed process~

Let me deduce the derivative of y = a Λ x with you!
f'(x)＝lim(△x→0)［f(△x＋x)－f(x)］/△x
＝lim(△x→0)［a∧(x＋△x)－a∧x］/△x
＝a∧xlim(△x→0)(a∧△x－1)/△x
＝a∧xlim(△x→0)(△xlna)/△x
＝a∧xlna.
That is: (a Λ x) '= a Λ xlna
In particular, when a = e,
(e∧x)'＝e∧x