Answer:
See verification below
Step-by-step explanation:
We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that
[tex]1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}[/tex]
Now, differentiate to obtain
[tex]\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}[/tex]
[tex]=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}[/tex]
To obtain the required form, extract a factor in both the numerator and denominator:
[tex]\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)[/tex]
