Why is rate of change of a fraction x/y the same as rate of change of x minus rate of change of y? Embarrassing question, I know, but I'm having some trouble demonstrating this to myself algebraically. I'm working my way through Solow's classic (1956) growth theory paper and am having trouble with the following logic (I adapted this from a set of slides explaining it). I'm not bad at math, but I couldn't seem to get the algebra to work, and I vaguely remember a teacher in high school explaining this kind of poorly. Can anyone show me why this is true?

Recall that r = K / P and so ∆r/r = ∆K/K - ∆P/P [where K is capital and P is population, making r the capital-labor ratio].

Below is how far I've gotten for a start. I've been pretty sick for almost a month and might just be making an incredibly basic error, but it seems like the rate of change ∆r/r is identical to what follows below, and I can't seem to manipulate what follows into ∆K/K - ∆P/P. Can anyone help? Thank you in advance!!

∆r/r = [(K2/P2 - (K1-P1)]/(K1/P1)

 A: Note that
$$r=\dfrac{K}{P}$$
$$\implies \log r =\log K-\log P$$
Differentiating both sides we get,
$$\dfrac{dr}{r}=\dfrac{dK}{K}-\dfrac{dP}{P}$$
You can also without logarithms, as follows.
$$RHS=\dfrac{P\Delta K-K\Delta P}{P^2\left(\dfrac{K}{P}\right)}$$
$$=\dfrac{\Delta r}{r}=LHS$$
(by division rule)
A: Note 
$$\Delta r=\frac{\partial r(K,P)}{\partial K}\Delta K+
\frac{\partial r(K,P)}{\partial P}\Delta P
=\frac1P \Delta K- \frac K{P^2} \Delta P $$
Divide both sides by $r=\frac KP$ to obtain
$$∆r/r = ∆K/K - ∆P/P$$
A: It is an approximation. For small $x$ we have ${1 \over 1+x} \approx 1-x$. Plus we ignore
any second order term.
You have ${\Delta R \over R}  = {P \over K} ({K+\Delta K \over P + \Delta P} -{ K \over P }) \approx {P \over K} ({K+\Delta K \over P }(1-{\Delta P \over P}) -{ K \over P }) = (1+ {\Delta K \over K})(1-{\Delta P \over P}) -1 \approx {\Delta K \over K} -{\Delta P \over P}$.
Elaboration:
The ${1 \over 1+x} \approx 1-x$ expression comes from the Taylor approximation
$f(y+h) \approx f(y)+f'(y)h$. In this case, $f(y) = { 1\over 1+y}$ and so
$f'(1) = -1$ and so ${1 \over 1+h} \approx 1 - 1 \cdot h$.
The above derivation shows how the approximation comes about without involving
derivatives directly.
More directly, you have a function $f(x) = {x_1 \over x_2}$ and you want to
see what the first order effects of a perturbation $h$ (which is two dimensional here) are. Again, Taylor
gives $f(x+h) \approx f(x) + Df(x)h$ and so $f(x+h)-f(x) \approx Df(x)h$.
Hence we have ${f(x+h)-f(x) \over f(x)} \approx {Df(x)h \over f(x)}$.
Since $Df(x)h = {x_2 h_1 - x_1 h_2 \over x_2^2}$, we have 
${Df(x)h \over f(x) } = {h_1 \over x_1} - {h_2 \over x_2}$.
Using the symbols in the question, we have
${\Delta R \over R} \approx {\Delta K \over K} -{\Delta P \over P}$.
Finally, note that the derivative of the function $l(x)= \log f(x)$ is
$Dl(x)h = {Df(x)h \over f(x)}$, so we have
$\log (R+\Delta R) - \log R \approx {\Delta R \over R}$.
A: From $r = \frac{K}{P}$, using your idea of $K_1$ & $P_1$ being the values before, with $K_2$ and $P_2$ being the values after, so their difference are the delta values, you get
$$\begin{equation}\begin{aligned}
\Delta r & = \frac{K_2}{P_2} - \frac{K_1}{P_1} \\
& = \frac{K_2 P_2 - K_1 P_1}{P_1 P_2} \\
& = \frac{K_2 P_1 - K_1 P_1 + K_1 P_1 - K_1 P_2}{P_1 P_2} \\
& = \frac{K_2 P_1 - K_1 P_1}{P_1 P_2} + \frac{K_1 P_1 - K_1 P_2}{P_1 P_2} \\
& = \frac{K_2 - K_1}{P_2} - \left(\frac{K_1}{P_2}\right)\left(\frac{P_2 - P_1}{P_1}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Now, to the first order of approximation, $\frac{1}{r} = \frac{P}{K} \approx \frac{P_2}{K_1}$. Thus, multiplying both sides of \eqref{eq1A} by this gives
$$\begin{equation}\begin{aligned}
\frac{\Delta r}{r} & \approx \frac{K_2 - K_1}{K_1} - \frac{P_2 - P_1}{P_1} \\
& \approx \frac{\Delta{K}}{K} - \frac{\Delta P}{P}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
