In a town of 50,000 people the number of people at time t who have influenza is $N(t)= (1000)/(1+9999e^{-t})$ where $t$ is measured in days. Note that the flu is spread by the one person who has it at $t=0$.
At what time $t$ is the rate of spreading the greatest?
I differentiate the function to find the rate of spreading which is $N'(t)= (99,990,000e^{-t})/(1+9999e^{-t})^2$.
Do I have to perform second derivative again and solve $N''(t)=0$ to find $t$ such that $N(t)$ is the greatest?
You have calculated the first derivative of $N$: $$ N'(t) = \frac{ae^{-t}}{(1+be^{-t})^2} $$ for some numbers $a$ and $b$ (omitted here for simplicity of presentation). What you need to do is to find the second derivative and equate it with 0: $$ N''(t) = \frac{-ae^{-t}(1+be^{-t})^2-ae^{-t}\cdot 2(1+be^{-t})\cdot -(be^{-t})}{(1+be^{-t})^4} = 0. $$ Because the denominator is non-zero, we only need to equate the enumerator with 0; namely, $$ (1+be^{-t}) = 2be^{-t} \Rightarrow be^{-t} = 1 \Rightarrow t = \ln(b). $$
In particular, with your calculation, $t=\ln(9999) \approx \ln(10^4) \approx 4\ln(10) \approx 9.2$.
