Context: Recently I have been studying Newton's Law of Cooling, we did a lab the other day and we are now trying to model The temperature of boiling water as it cools down. We collected data every 10 seconds up to 20 minutes. When it came to modelling the function, we simply integrated Newton's Law of Cooling, however, when we were solving for the relative decay rate of the model ( the k in $e^{kx}$) We found that the relative decay rate of the model was not constant, that it changed based on the Temperature and the time. So in an attempt to solve it, I was wondering, is the relative growth rate the same as the derivative, or in other words, is it correct if I were to try modelling this data using a function of the form, $$T(t)=Ae^{(r(T_{room}-T))x}$$ Or is in fact the derivative different from the relative growth rate of the function? Or in other words does $k=\frac{dT}{dt}$?
Edit: For added context on how I calculated k. I used the original formula, I have yet to test the mentioned example above. Data is linked here:https://www.desmos.com/calculator/mqpnomvey1. Given a point such as $(10, 84.4995168)$, Using substitution into a basic integrated model, $$T(t)=(T_{0}-T_{room})e^{-kt}+T_{room}$$ Room temperature is 25 Degrees $$T(t)=(85.125-25)e^{-kt}+25$$ $$84.4995168=60.125e^{-10k}+25$$ $$k=-\frac{1}{10}\ln(\frac{84.4995168-25}{60.125})$$ This was repeated for every data point using excel.