Why $\frac{1}{m}\frac{dm}{dt}= \frac{d}{dt}\left ( \log_{e}m \right )$ is true?

Question

In the explanation of relative growth rate calculation, in chapter 2 of R.A. Fisher's Book Statistical Methods for Research Workers, it is shown the following equality:

$$\frac{1}{m}\frac{dm}{dt}= \frac{d}{dt}\left ( \log_{e}m \right )$$

I could not see why this is true, since $(1/x)' = -1/x^{2}$ and $(\ln x)'= \frac{1}{x}$

Do I am missing something? why the equality is true?

Answer 1

Here, $m$ is some function of $t$.
By Chain rule: $(fog) '(t) =f' (g(t)) g'(t) $.
In your case, $f(t) =\ln|t|, g(t) =m \implies g'(t) =dm/dt $ and $f'(t) =1/t $ for all $t$ except $t=0$. Hence, $(fog) '(t) = f' (g(t)) g'(t) =\frac{1}{g(t) } \frac{dm} {dt} =\frac{1} {m}\frac{dm} {dt}$.

Answer 2

Note $$\frac{d}{dt} (\ln m) = \frac{d}{dm} (\ln m )\times \frac{dm}{dt} = \frac 1m \frac{dm}{dt}$$