I want to analyse a logarithmic growth curve in more detail. Especially I would like to kow the time point when the slope becomes >0 (which is the starting point of growth after a lag phase). Therefore I fitted a logarithmic function to my growth data with the grofit package of R. I got values for the three parameters (lambda, mu, maximal assymptote). Now I thought, I could use the first derivative of the logarithmic growth function to put mu=0 (the slope of any time point during growth) and this way solve the equation for the time (x). I'm not sure if this is possible, since the mu=0 will be correct for a longer timespan at the beginning of the curve (and no unique timepoint). But maybe I could approximate to that point by putting mu=0.01. This should be more specific. Anyway I used the Deriv package to find the first derivative of my logarithmic function:
Deriv(a/(1+exp(((4*b)/a)*(c-x)+2)), "x")
where a=assymptote, b=maximal slope, c=lambda.
As a result I got:
{.e2 <- exp(2 + 4 * (b * (c - x)/a))
4 * (.e2 * b/(.e2 + 1)^2)}
Or in normal writing:
f'(x)=(4*exp(2+((4b(c-x))/a))*b)/((exp(2+((4b(c-x))/a))+1)^2)
Now I would like to solve this function for x with f'(x)=0.01. Can anyone tell me, how best to do it?
Also, do you have comments on my way of thinking or the R functions I used?
Thank you. Anne
Thank you very much for your efforts. Unfortunately, none of the above solutions worked for me :-( I figured the problem out the old fashioned way (pencil + paper + mathematics book). Have a good day Anne