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I'm struggling with simplifying the derivative of the function.

Here is the function and derivative:

$$\frac{dP}{dt}=P(1-P)\\P=\frac{c_1e^t}{1+c_1e^t}$$

I have to get the function to "look" like the derivative right? Ignore the arrow in my work. Thanks for taking a look at my problem. enter image description here

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    Your final result is incorrect. In the last line, when you factored out $c_{1}e^{t}/(1+c_{1}e^{t})$ from the previous line, you should still have a $1+c_{1}e^{t}$ term in the numerator, not a $1$. – 2017-01-27

1 Answers 1

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Step 1:

$$P = \dfrac{c_1 e^t}{1 + c_1 e^t} \implies \dfrac{dP}{dt} = \dfrac{c_1 e^t}{c_1 e^t+1}-\dfrac{c_1^2 e^{2 t}}{\left(c_1 e^t+1\right)^2} = \dfrac{c_1 e^t}{\left(c_1 e^t+1\right)^2}$$

Step 2:

$$\dfrac{dP}{dt} = P(1-P) = \dfrac{c_1 e^t}{1 + c_1 e^t}\left(1 - \dfrac{c_1 e^t}{1 + c_1 e^t}\right) = \dfrac{c_1 e^t}{c_1 e^t+1}-\dfrac{c_1^2 e^{2 t}}{\left(c_1 e^t+1\right)^2} = \dfrac{c_1 e^t}{\left(c_1 e^t+1\right)^2}$$