logarithm transition for the population growth equation

Question

Analyzing the growth population equation I came across with the below transition

$$\frac{d \ lnN}{dt}=\frac{dN}{N dt}$$

which I don't quite follow. Can anybody clarify this please?

Answer 1

Note that from the chain rule we have

$$\begin{align} \frac{d\log(N)}{dt}&=\frac{d\log(N)}{dN}\times \frac{dN}{dt}\\\\ &=\frac{1}{N}\frac{dN}{dt} \end{align}$$

Answer 2

Notice for the LHS, by the chain rule:

$$\frac{\text{d}}{\text{d}t}\left(\ln\left(\text{N}(t)\right)\right)=\frac{\frac{\text{d}}{\text{d}t}\left(\text{N}(t)\right)}{\text{N}(t)}=\frac{\text{N}'(t)}{\text{N}(t)}$$

And for the RHS:

$$\frac{1}{\text{N}(t)}\cdot\frac{\text{d}}{\text{d}t}\left(\text{N}(t)\right)=\frac{1}{\text{N}(t)}\cdot\text{N}'(t)=\frac{\text{N}'(t)}{\text{N}(t)}$$

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For the chain rule:

If a function $g(x)$ is differentiable at the point $x$ and a function $f(x)$ is differentiable at the point $g(x)$, then the composition $f\circ g$ is differentiable at $x$. Furthermore, let $y=f(g(x))$ and $u=g(x)$, then $\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}\cdot\frac{\text{d}u}{\text{d}x}$.