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

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.

• d(PY) = Y dP + P dY is the exact definition of the total derivative of f(P,Y) = PY. – dsmithecon Mar 8 '18 at 21:46

## 2 Answers

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.

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.

• Thanks, your answer does help. However, I've actually found a different explanation by the author, but I can't seem to follow it. If it's not too much trouble, would you be so kind as to clear up the question I've added to my original post? Thanks very much. – Sasaki Feb 27 '18 at 15:18
• He is taking the total derivative in that example. en.wikipedia.org/wiki/Total_derivative so he takes the derivative with respect to P and Y. That lets you know how much a function moves if you change both P and Y. I don't think that is what he is doing here because he'd have to divide the left side by MV and the right by PY to convert the total derivative to percent changes. There should be some equivalence between the two approaches, but I don't immediately see it. – dsmithecon Feb 27 '18 at 15:49
• From what I understood, when you take the total derivative of $f(x,y)$, you're still differentiating with respect to something, the difference being that in this case you treat $x$ as a function of $y$ or vice-versa or both $x$ and $y$ as functions of an exogenous argument. But thanks anyway. – Sasaki Mar 6 '18 at 20:35
• The total derivative of f(x,y) is the sum of the derivative with respect to x and the derivative with respect to y (times how much x and y are changing). That is what differentiates the total derivative from the derivative with respect to only one argument. When you do it there is no notion of whether variables of the function are exogenous or endogenous. – dsmithecon Mar 7 '18 at 18:07