Multivariate implicit differentiation (Samuelson, 1948)

Question

I am studying "The Simple Mathematics of Income Determination", by Paul Samuelson (from book is called "Macroeconomia (artigos selecionados)", by APEC-CAEN. It is the 1st chapter of the book, the 13th page), and I am having some troubles in understanding a calculus passage.

First, Samuelson states the income equation:

$(1)$ $ Y = C (Y - \bar{W}) + \bar{I} + \bar{G} $

Where $Y$ corresponds to income, $C$ corresponds to consumption, $\bar{W}$ corresponds to taxes (actually, I don't exactly know the odd reason why the letter W was picked as taxes instead of a more common letter as T), $\bar{I}$ corresponds to investment and $\bar{G}$ corresponds to government spending). For purpose of simplification, these three last variables are treated as constants.

Then, $Y$ is differentiatied in $\bar{G}$ (which I suppose it is an implicit differentiation, in the form of $\frac{dy}{dx} = - \frac{F_x}{F_y}$):

$(2)$ $\frac{dY}{d\bar{G}} = \frac{1}{1 - C'(Y - \bar{W})}$

But my doubt emerges in the following passage:

$(3)$ $\frac{dY}{d\bar{(-W)}} = \frac{C'(Y - \bar{W})}{1 - C'(Y - \bar{W})} = \frac{dY}{d\bar{G}} - 1 $

I realize that Samuelson did the same implicit differentiation as in step 2 and I can visualize that

$\frac{C'(Y - \bar{W})}{1 - C'(Y - \bar{W})} = \frac{1}{1 - C'(Y - \bar{W})} * C'(Y - \bar{W}) = \frac{dY}{d\bar{G}} * C'(Y - \bar{W})$

but, how does $\frac{dY}{d\bar{G}} * C'(Y - \bar{W})$ turns to $\frac{dY}{d\bar{G}} - 1 $? I mean, how did this $-1$ replace the multiplying term $C'(Y - \bar{W})$

I tried to get it, but unsuccessfully. I hope you could shed some light on it.

Answer 1

We start from the equation: $$ Y = C(Y - \overline{W}) + \overline{I} + \overline{G} $$ First, we totally differentiate with respect to $(-\overline{W})$ (taking into account that $Y$ is a function of $\overline{W}$, $\overline{I}$ and $\overline{G}$) $$ dY = C'(Y - \overline{W})dY + C'(Y - \overline{W}) d(-\overline{W}). $$ Bringing the $dY$ terms on one side gives: $$ (1 - C'(Y - \overline{W}))dY = C'(Y - \overline{W}) d(-\overline{W}),\\ \to \frac{dY}{d(-\overline{W})} = \frac{C'(Y - \overline{W})}{1 - C'(Y - \overline{W})} $$ Next, we totally differentiate with respect to $\overline{G}$: $$ dY = C'(Y - \overline{W})dY + d \overline{G} $$ Again gathering terms gives: $$ (1 - C'(Y - \overline{W})) dY = d \overline{G}\\ \to \frac{dY}{d \overline{G}} = \frac{1}{1 - C'(Y - \overline{W})} $$ Therefore: $$ \begin{align*} \frac{dY}{d \overline{G}} - 1 &= \frac{1}{1 - C'(Y - \overline{W})} - 1,\\ &=\frac{1 - 1 +C'(Y - \overline{W})}{1 - C'(Y - \overline{W})},\\ & = \frac{C'(Y - \overline{W})}{1 - C'(Y - \overline{W})},\\ &= \frac{dY}{d(- \overline{W})} \end{align*} $$