UndIch

Anderer Vorschlag

$f'(t)={\frac {e^{t}-e^{2t}}{(1+e^{t})^{3}}}$

--Physics 14:59, 13. Apr. 2007 (CEST)

Also: $f'(t)={\frac {e^{t}*(1+e^{t})^{2}-e^{t}*2e^{t}(1+e)}{(1+e)^{4}}}={\frac {e^{t}(1+e^{t})-2e^{2}t}{(1+e^{t})^{3}}}={\frac {e^{t}-e^{2t}}{(1+e^{t})^{3}}}$

$f(x)=a*e^{b*x}$

Gegeben ist:

$f(0,27)=2f(0,74)=4$

$a*e^{0,27b}=2a*e^{0,74b}=4$

a*e^(0,27x)=2 a=\frac{4}{e^{0,74b}}

\frac{4}{e^{0,74x}}*a*e^{0,27b}=2 a*e^{0,74b)=4

e^{-0,47t}=2 a*e^{0,74b)=4

b=\frac{2}{ln(4)} a*e^{0,74x)=4

b=1,474 a*e^{0,74x)=4

</math> ich denke weiter weist du