[Risolto] Derivate

y'=dy/dx=-k/[f(x)]^2*[f'(x)]

Applicando la regola segnata sopra:

- 100/(1 + 2·e^(- 2·x))^2·(- 4·e^(- 2·x))=

=400·e^(- 2·x)/(1 + 2·e^(- 2·x))^2=

=400·e^(- 2·x)/(1 + 2/e^(2·x))^2=

=400·e^(- 2·x)/(e^(-4x)(e^(2x) + 2)^2)=

=400·e^(2·x)/(e^(2·x) + 2)^2

* f(x) = 100/(1 + 2*e^(- 2*x)) = 100*e^(2*x)/(e^(2*x) + 2) ≡ g(u) = N(u)/D(u) = 100*e^u/(e^u + 2)
* f'(x) = df/dx = 400*e^(2*x)/(e^(2*x) + 2)^2
Infatti
* du/dx = 2
* df/dx = (dg/du)*du/dx = 2*(dg/du) = 200*(d/du e^u/(e^u + 2))
* d/du N(u) = e^u
* d/du D(u) = e^u
* d/du N(u)/D(u) = (D*N' - N*D')/D^2
Tu rimonti i pezzi e hai dimostrato il risultato.

é una logistica

la derivata   é 100 *   - g'(x)/g^2(x)   con g(x) = (1 + 2e^(-2x))

100 / ( 1 + 2e^-2x)

y' = 100 * (- 2*e^(-2x))*(-2) /( 1 + 2 e^(-2x))^2 = 400 e^(-2x)/(1 + 2 e^(-2x)) =

= 400 e^(2x)/[ e^(4x) * (1 + 2e^(-2x))^2 ] = 400 e^(2x)/(e^(2x) + 1)^2