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

CC BY-SA 4.0
2021-05-23T15:47:39.683Z
Thank you, tdm, that was exactly what I needed!
PGabriel96
CC BY-SA 4.0
2021-05-23T09:01:01.250Z
2021-05-23T09:01:01.250Z
tdm
<p>I am studying &quot;The Simple Mathematics of Income Determination&quot;, by Paul Samuelson (from book is called &quot;Macroeconomia (artigos selecionados)&quot;, 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.</p>
<p>First, Samuelson states the income equation:</p>
<p><span class="math-container">$(1)$</span> <span class="math-container">$ Y =  C (Y - \bar{W}) + \bar{I} + \bar{G} $</span></p>
<p>Where <span class="math-container">$Y$</span> corresponds to income, <span class="math-container">$C$</span> corresponds to consumption, <span class="math-container">$\bar{W}$</span> corresponds to taxes (actually, I don't exactly know the odd reason why the letter W was picked as taxes instead of a more common letter as T), <span class="math-container">$\bar{I}$</span> corresponds to investment and <span class="math-container">$\bar{G}$</span> corresponds to government spending). For purpose of simplification, these three last variables are treated as constants.</p>
<p>Then, <span class="math-container">$Y$</span> is differentiatied in <span class="math-container">$\bar{G}$</span> (which I suppose it is an implicit differentiation, in the form of <span class="math-container">$\frac{dy}{dx} = - \frac{F_x}{F_y}$</span>):</p>
<p><span class="math-container">$(2)$</span> <span class="math-container">$\frac{dY}{d\bar{G}} = \frac{1}{1 - C'(Y - \bar{W})}$</span></p>
<p>But my doubt emerges in the following passage:</p>
<p><span class="math-container">$(3)$</span> <span class="math-container">$\frac{dY}{d\bar{(-W)}} = \frac{C'(Y - \bar{W})}{1 - C'(Y - \bar{W})} = \frac{dY}{d\bar{G}} - 1  $</span></p>
<p>I realize that Samuelson did the same implicit differentiation as in step 2 and I can visualize that</p>
<p><span class="math-container">$\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})$</span></p>
<p>but, how does <span class="math-container">$\frac{dY}{d\bar{G}} *  C'(Y - \bar{W})$</span> turns to <span class="math-container">$\frac{dY}{d\bar{G}} - 1  $</span>?
I mean, how did this <span class="math-container">$-1$</span> replace the multiplying term <span class="math-container">$C'(Y - \bar{W})$</span></p>
<p>I tried to get it, but unsuccessfully. I hope you could shed some light on it.</p>

CC BY-SA 4.0
2021-05-22T20:05:22.807Z
By the way, there is another pertinent information about it. In the following paragraph after the step 3, it is said that "in words, a reduction of taxes and a increase in government spending increase income, but the taxes reduction effect on income is smaller than the government spending effect on income by 1". It certainly clarifies the meaning of "dY/dG - 1", but I still cannot see it mathematically.
PGabriel96
<macroeconomics><mathematical-economics><national-income>
Multivariate implicit differentiation (Samuelson, 1948)
PGabriel96
# Multivariate implicit differentiation (Samuelson, 1948)

(Asked by: username_0 on 2021-05-22)

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.

**Tags:** macroeconomics, mathematical-economics, national-income
> **Comments:**
> **username_0**: By the way, there is another pertinent information about it. In the following paragraph after the step 3, it is said that "in words, a reduction of taxes and a increase in government spending increase income, but the taxes reduction effect on income is smaller than the government spending effect on income by 1". It certainly clarifies the meaning of "dY/dG - 1", but I still cannot see it mathematically.  

---

## 1 answers

### Answer by username_1 on 2021-05-23. Score: 4 | Accepted answer

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:
$$
\begin{align\*}
\frac{dY}{d \overline{G}} - 1 &= \frac{1}{1 - C'(Y - \overline{W})} - 1,\\
&=\frac{1 - 1 +C'(Y - \overline{W})}{1 - C'(Y - \overline{W})},\\
& = \frac{C'(Y - \overline{W})}{1 - C'(Y - \overline{W})},\\
&= \frac{dY}{d(- \overline{W})}
\end{align\*}
$$

> **Comments:**
> **username_0**: Thank you, tdm, that was exactly what I needed!  

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