How to get the percentage-change form of the quantity equation?

Question

In my book, the quantity equation is presented as $$M\bar{V}=PY$$ And on the next page, the author presents its percentage-change form: $$\% \ \text{Change in}\ M + \% \ \text{Change in}\ V = \% \ \text{Change in}\ P + \% \ \text{Change in}\ Y$$ My question is: where exactly does the second equation come from? I've tried manipulating the original equation to see if I could get the percentage-change form, but I didn't get anywhere. Any help will be appreciated. Thanks.

@edit

I've actually found an explanation in a footnote in another chapter of the book (see below), but there's still one thing I don't understand: what are we differentiating with respect to? If it's $Y$, then $d(PY)=P$; if it's $P$, then $d(PY)=Y$; unless we're actually differentiating with respect to some third variable of which $P$ and $Y$ are both functions. In this case, what would that variable be? (I may be misunderstanding it too). Thanks.

Here's the explanation (Mankiw Macroeconomics 7th edition, page 26):

The proof that this trick works begins with the product rule from calculus: $$d(PY)=Y \ dP+P \ dY$$ Now divide both sides of this equation by PY to obtain: $$d(PY)/PY)=dP/P+dY/Y$$ Notice that all three terms in this equation are percentage changes.

Answer 1

In short the answer is: time - all variables depend on time and what you call "% change in M" is better called "growth rate"

For simplicity let's consider only one the right side of the quantity equation and donate this side with $Z=Y P$:

The growth rate of $Z$ equals ${dZ\over dt} {1\over Z} = {\dot Z \over Z}$ with $dZ \over dt $ as the total derivative of $Z$ with respect to time $t$.

Let's do some math with the above formula for $Z$ where $Z$ depends on $Y$ and $P$ which both again depend on $t$:

${dZ\over dt} {1\over Z}$ = $({dZ \over dY} {dY \over dt}+{dZ \over dP} {dP \over dt}) {1 \over Z }$

${dZ\over dt} {1\over Z}$ = $({dYP \over dY} {dY \over dt}+{dYP \over dP} {dP \over dt}) {1 \over YP }$

${dZ\over dt} {1\over Z}$ = $({P} {dY \over dt}+{Y} {dP \over dt}) {1 \over YP }$

${dZ\over dt} {1\over Z}$ = $({P} {dY \over dt}+{Y} {dP \over dt}) {1 \over YP }$

${dZ\over dt} {1\over Z}$ = ${dY \over dt} {1 \over Y} +{dP \over dt} {1 \over P}$

Which is the same as the sum of the growth rates with ${\dot Z \over Z}={\dot Y \over Y}+{\dot P \over P}$ which is for small values pretty close to taking logs which was already pointed out.

The same argument applys the the left side of the euqation.

Answer 2

I think the author is just taking logs of the original equation. The natural log approximates percent changes for small changes. More info here. Using that and properties of the log, $\ln(ab)=\ln(a)+\ln(b)$, you get that formula.