Why is rate of change of a fraction x/y the same as rate of change of x minus rate of change of y?

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)

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

$$∆r/r = ∆K/K - ∆P/P$$

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 $\log (R+\Delta R) - \log R \approx {\Delta R \over R}$.