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August 6th, 2014, 03:02 AM   #1
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Gompertz equation - differential equation

Hello everyone!

How do you get from the first equation to the second (see uploaded image)?
Is the second equation the first derivative of the differential equation? In this case, how would you thus find the second derivative of this differential equation?

Thank you very much!!
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 August 6th, 2014, 03:24 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus You should have in mind that N is changing with respect to t, so it is N(t). $\displaystyle \frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\cdot\frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\left[N'(t)\ln\left(\frac{K}{N(t)}\right)+N(t)\ln'\left( \frac{K}{N(t)}\right)\right]$ Thanks from Sonprelis
August 6th, 2014, 03:35 AM   #3
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Quote:
 Originally Posted by ZardoZ You should have in mind that N is changing with respect to t, so it is N(t). $\displaystyle \frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\cdot\frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\left[N'(t)\ln\left(\frac{K}{N(t)}\right)+N(t)\ln'\left( \frac{K}{N(t)}\right)\right]$

Thank you very much for your answer, I am still somewhat confused though. I've never really studied differential equations and have chosen to do so for a mathematical exploration. Could you thus please expand on the explanation? Why and how is it that it goes from one form to the other?
I would be infinitely grateful for your help!

 August 6th, 2014, 03:37 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Do you know anything about differentiation rules? $\displaystyle \left(f(t)g(t)\right)'=f'(t)g(t)+f(t)g'(t)$ in our case $f(t)=N(t)$ and $g(t)=\ln\left(\frac{K}{N(t)}\right)$, the ' implies $\frac{d}{dt}$. Thanks from Sonprelis Last edited by ZardoZ; August 6th, 2014 at 03:40 AM.
 August 6th, 2014, 09:50 AM #5 Newbie   Joined: Apr 2014 From: Italy Posts: 27 Thanks: 0 Now it's clear, thank you However, I still wanted to know... when we apply the d/ dt to the equation rN ln(K/N), what is it exactly that we're doing? Are we finding the first derivative of the equation? Can this be done with a differential equation? The original equation is indeed a differential one: dN / dt = rN ln (K/N) and then we have to apply the d/dt to it: d/dt [rN ln (K/N)]
 August 6th, 2014, 10:02 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus first derivative. Rate of change with respect to t (probably time here)! Thanks from Sonprelis
August 6th, 2014, 10:07 AM   #7
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Quote:
 Originally Posted by ZardoZ Rate of change with respect to t (probably time here)!

...and how would the second derivative look like for the following function?

dN / dt = rN ln (K/N)

The first derivative we now know to be:

r[dN/dt ln(K/N) + N d/dt ln(K/N)]

Thank you very much for your kind help!!

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