they used compound interest formula for doing constant rate of population growth. the gave this answer for the differential equation. i want to understand what they did/ how they did it.
A differential equation including initial conditions.
$$P = P(0) (1+r)^t$$
$$\frac{dP}{dt} = P(0) \cdot \ln{(1+r)} \cdot (1+r)^t$$
$$\frac{dP}{dt} = k (1+r)^t$$
can someone explain what they did each step I feel like I'm missing something
On
Given $P = P_o (1+r)^t$
I'll apply natural logarithm on both sides of the equation
$\ln P = \ln (P_o (1+r)^t)$
$\ln P = \ln P_o + t\ln (1+r)$
Then I'll differentiate the equation both sides
$\frac{dP}{P} = 0 + \ln (1+r) dt$
$\frac{dP}{P} = \ln (1+r) dt$
$\frac{dP}{dt} = P\ln (1+r)$
but $P = P_o (1+r)^t$
thus $\frac{dP}{dt} = P_o (1+r)^t\ln (1+r)$
$\frac{dP}{dt} = P_o\ln (1+r) (1+r)^t$
but $P_o\ln (1+r) = k$ where by $k$ is constant
therefore $\frac{dP}{dt} = k (1+r)^t$
\begin{align*} \frac d{dt} a^t = {} & \lim_{h\to0} \frac{a^{t+h} - a^t} h \\[2pt] & \text{(definition of “derivative”)} \\[10pt] = {} & \lim_{h\to0}\left( a^t\cdot \frac {a^h-1} h \right) \\[10pt] = {} & a^t\cdot\lim_{h\to0} \frac{a^h -1 } h \\[2pt] & \text{since $a^t$ does not depend on $h$} \\[10pt] = {} & \big(a^t \times\text{constant}\big) \\ & \text{since the limit does not depend on $t$} \end{align*}
Just what the “constant” is can be seen via the chain rule if one knows that the constant is $1$ when $a=e\approx 2.71828\ldots\,$:
\begin{align} \frac d {dt} a^t = {} & \frac d {dt} e^{t\ln a} = e^{t\ln a}\cdot \frac d {dt} (t\ln a) \\[10pt] = {} & a^t \ln a. \end{align}
And here you see what is “natural” about $e$: It is the base for which this “constant” is $1.$