Can someone show me the steps involved in computing the derivative of the Gompertz trajectory function for $N(t)$?
$$ N(t) = K \cdot \exp\Big( e^{-\theta \cdot t} \cdot \ln \big(\frac{N_0}{K}\big)\Big) $$
$K$ is the carrying capacity, $N_0$ is the initial population size, and $\theta$ is the growth rate. I know the derivative is $ \frac{dN(t)}{dt} = \theta \cdot N(t) \cdot \ln \frac{K}{N(t)} $ from the solution being presented in research articles but it would help to see how it is derived step-by-step. Thank you.
The function simplifies to $$N(t) = K\cdot (\frac{K}{N_0})^{e^{-\theta t}}$$
Taking the derivative with respect to $t$, using $\frac{d}{dt} a^{f(t)}=a^{f(t)}\cdot \ln (a)\cdot \frac{df}{dt}$, we obtain
$$\frac{dN}{dt}=K\cdot (\frac{K}{N_0})^{e^{-\theta t}}\cdot \ln{(\frac{K}{N_0})} \cdot\frac{d}{dt}e^{-\theta t}$$
Replacing the first to terms in the product by $N(t)$ and computing the derivative, we obtain
$$\frac{dN}{dt} = -\theta e^{-\theta t}N(t)\ln (\frac{K}{N_0})=-\theta N(t)\ln{(\frac{K}{N_0})}e^{-\theta t}$$
From the given form of $N(t)$, we see that $-\ln{(\frac{K}{N_0})}e^{-\theta t}=\ln (\frac{K}{N(t)})$, so we can now write
$$\frac{dN}{dt} = \theta N(t) \ln (\frac{K}{N(t)})$$
There may be simpler ways to go about this but I just went along with whatever came to mind, sorry if I'm missing a shortcut.