<p>Total differentiation means you are differentiating all variables in the expression. So if we start with</p>
<p><span class="math-container">$$y = C\{[y + B - T(y + B)]; [M + \frac{B}{r}]\} + I(r) + G$$</span></p>
<p>we get:</p>
<p><span class="math-container">$$ d y = C_y^{'} dy + C_y^{'} dB - C_y^{'} dTdy - C_y^{'} dTdB + C_A^{'} dM + C_A^{'} \left( \frac{dBr-Bdr}{r^2} \right) + I'dr + dG $$</span></p>
<p>Here you are using just basic calculus such as chain rule accordding to which <span class="math-container">$\frac{d}{dx}[y(u(x))] = \frac{dy}{du} \frac{du}{dy}$</span>, plus you use the rule that with total differentiation  <span class="math-container">$dy(x_1,x_2) = y_1' dx_1 +y_2'dx_2$</span> and also quotient rule <span class="math-container">$\frac{d}{dx}[F(x)/G(x)]= \frac{F'G(x) - F(x)G'}{G(x)^2}$</span>.</p>
<p>Next you just move everything to the LHS of the equation to get</p>
<p><span class="math-container">$$dy - C_y^{'} dy - C_y^{'} dB + C_y^{'} T'dy + C_y^{'} T'dB - C_A^{'} dM - C_A^{'} \left( \frac{dBr-Bdr}{r^2} \right) - I'dr - dG = 0$$</span></p>
<p>Now just collect terms and clean up the mess above and you directly get:</p>
<p><span class="math-container">$$(1 + C_y^{'}(1 -  T'))dy - (I_r - C_A \frac{B}{r^2})dr - [C_y (1 - T') + \frac{C_A}{r}]dB- C_A^{'} dM  - dG = 0$$</span></p>
<p>The expression looks difficult because it has a lot of terms but you should not get intimidated by its look.</p>

CC BY-SA 4.0
2021-04-22T23:07:33.170Z
That kind of clarification was exactly what I needed, thank you, muflon. By the way, I consulted my book and the "dG" term is written as a positive one, do you think it can be a mistake in the book? The way you developed the total differentiation here with the negative "dG" seems a lot more reasonable for me. Thanks again.
PedroColangelo
CC BY-SA 4.0
2021-04-22T23:12:22.107Z
@PedroColangelo you are welcome, if you think this answer answered your Q consider accepting it. I think that must have been a typo because except for moving it to other side we do not do any operation with dG so it should be negative
1muflon1
CC BY-SA 4.0
2021-04-22T20:47:23.380Z
2021-04-22T20:47:23.380Z
1muflon1
<p>everyone. I'm having a hard time trying to figure out how the total differentiation occurs in this exercise.</p>
<p>In this excerpt, from Inflação e Crises&quot;, by Affonso Celso Pastore, the author develops an alternative form of the IS curve. First, he states some equations for its development.</p>
<p>The aggregate demand as the sum of household consumption (C), investment (I) and government spending (G):</p>
<p><span class="math-container">$(1)$</span>     <span class="math-container">$y = C + I + G$</span></p>
<p>Consumption as a function of disposable income (<span class="math-container">$y_d$</span>) and non-human wealth (A):</p>
<p><span class="math-container">$(2)$</span> <span class="math-container">$C = C(y_d, A)$</span>      with <span class="math-container">$0 &lt; C_y &lt; 1$</span>; <span class="math-container">$C_a &gt; 0$</span></p>
<p>Investment as a function of the real interest rate:</p>
<p><span class="math-container">$(3)$</span> <span class="math-container">$I = I(r)$</span> with <span class="math-container">$I_r &lt; 0$</span></p>
<p>Non-human wealth (A) being equal to the sum of the money stock and the income originated by the stock of government bonds (B) divided by the real interest rate:</p>
<p><span class="math-container">$(4)$</span> <span class="math-container">$A = M + \frac{B}{r}$</span></p>
<p>Disposable income (y_d ) being equal to the sum of the labor income (y) and the income originated by the stock of government bonds (B) minus the taxation, that applies both to the labor income and the income originated by the stock of government bonds:</p>
<p><span class="math-container">$(5)$</span> <span class="math-container">$y_d = y + B - T (y + B)$</span></p>
<p>On (5), the author also states that</p>
<p><span class="math-container">$T = T (y + B)$</span> with <span class="math-container">$T'&gt; 0$</span></p>
<p>With this definitions, we finally arrive at this form of the IS curve:</p>
<p><span class="math-container">$(6)$</span> <span class="math-container">$y = C\{[y + B - T(y + B)]; [M + \frac{B}{r}]\} + I(r) + G$</span></p>
<p>And we finally get where I'm having troubles in understanding it, when he uses the total differentiation:</p>
<p><span class="math-container">$(7)$</span> <span class="math-container">$[1 - C_y(1 - T')]dy - (I_r - C_A \frac{B}{r^2})dr - [C_y (1 - T') + \frac{C_A}{r}]dB - C_AdM + dG = 0$</span></p>
<p>I've already tried so hard to understand it, but it always seems that I am missing some parts of the process. I explicitly have three doubts:</p>
<p>(i) I would like to know if there is a particular reason that, in this total differentiation, we use the derivatives dy, dr, dB, dM and dG, but we do not use dT, since T is a variable that appears in the original equation too.</p>
<p>(ii) Second doubt: although I can see how the &quot;inner&quot; derivation processes occur, I am not sure I can correctly interpret the signs (the &quot;plusses&quot; and the &quot;minuses&quot;). I mean, why C_a dM is a negative term while dG is a positive term? This doubt extend to the other terms as well.</p>
<p>(iii) The third one: I have a doubt in the derivation of the T variable in dy and dB. I don't know exactly how it got derived in dy and dB.</p>
<p>By the way, if necessary, I can provide some more informations of my book for contextualization purposes. Sorry if I'm sounding confused, I study economics by self-taught and this is my first time using Economics Stack Exchange.</p>
<p>Thanks for your attention!</p>

CC BY-SA 4.0
2021-04-22T18:59:17.103Z
Of course. I did not mention it before because it is not from any of the major economics textbooks and I thought that it would not be that helpful. It is "Inflação e Crises", by Affonso Celso Pastore.
PedroColangelo
CC BY-SA 4.0
2021-04-22T19:02:57.457Z
What is it that you do not understand? Do you know what total differentation is?
Giskard
CC BY-SA 4.0
2021-04-22T19:36:17.293Z
I have a basic knowledge on total differentiation cause I'm studying economics by self-taught. I mean, I explicitly have 3 doubts about it that I thought that if someone explained properly this total differentiation, I could understand it. I will write in the comments these three and edit the post for more clarity. Sorry if Im sounding confused, it's been 7 days that I'm trying to understand it completely and also it's my first time using StackExchange.
PedroColangelo
CC BY-SA 4.0
2021-04-22T19:41:39.220Z
First doubt: I would like to know if there is a particular reason that, in this total differentiation, we use the derivatives dy, dr, dB, dM and dG, but we do not use dT, since T is a variable that appears in the original equation.

Second doubt: although I can see how the "inner" derivation processes occur, I am not sure I can correctly interpret the signs (the "plusses" and the "minuses"). I mean, why C_a dM is a negative term while dG is a positive term? This doubt extend to the other variables as well.
PedroColangelo
CC BY-SA 4.0
2021-04-22T19:45:30.880Z
The third one: I have a doubt in the derivation of the T variable in dy and dB. I don't know exactly how it got derived in dy and dB.
PedroColangelo
CC BY-SA 4.0
2021-04-22T20:06:37.763Z
@PedroColangelo I added that reference to the post. Whenever you mention some sources you should refer to them even if they are not widely known
1muflon1
<macroeconomics><mathematical-economics>
Total differentiation in a IS curve context
PedroColangelo
# Total differentiation in a IS curve context

(Asked by: username_0 on 2021-04-22)

everyone. I'm having a hard time trying to figure out how the total differentiation occurs in this exercise.

In this excerpt, from Inflação e Crises", by Affonso Celso Pastore, the author develops an alternative form of the IS curve. First, he states some equations for its development.

The aggregate demand as the sum of household consumption (C), investment (I) and government spending (G):

$(1)$ $y = C + I + G$

Consumption as a function of disposable income ($y\_d$) and non-human wealth (A):

$(2)$ $C = C(y\_d, A)$ with $0 < C\_y < 1$; $C\_a > 0$

Investment as a function of the real interest rate:

$(3)$ $I = I(r)$ with $I\_r < 0$

Non-human wealth (A) being equal to the sum of the money stock and the income originated by the stock of government bonds (B) divided by the real interest rate:

$(4)$ $A = M + \frac{B}{r}$

Disposable income (y\_d ) being equal to the sum of the labor income (y) and the income originated by the stock of government bonds (B) minus the taxation, that applies both to the labor income and the income originated by the stock of government bonds:

$(5)$ $y\_d = y + B - T (y + B)$

On (5), the author also states that

$T = T (y + B)$ with $T'> 0$

With this definitions, we finally arrive at this form of the IS curve:

$(6)$ $y = C\{[y + B - T(y + B)]; [M + \frac{B}{r}]\} + I(r) + G$

And we finally get where I'm having troubles in understanding it, when he uses the total differentiation:

$(7)$ $[1 - C\_y(1 - T')]dy - (I\_r - C\_A \frac{B}{r^2})dr - [C\_y (1 - T') + \frac{C\_A}{r}]dB - C\_AdM + dG = 0$

I've already tried so hard to understand it, but it always seems that I am missing some parts of the process. I explicitly have three doubts:

(i) I would like to know if there is a particular reason that, in this total differentiation, we use the derivatives dy, dr, dB, dM and dG, but we do not use dT, since T is a variable that appears in the original equation too.

(ii) Second doubt: although I can see how the "inner" derivation processes occur, I am not sure I can correctly interpret the signs (the "plusses" and the "minuses"). I mean, why C\_a dM is a negative term while dG is a positive term? This doubt extend to the other terms as well.

(iii) The third one: I have a doubt in the derivation of the T variable in dy and dB. I don't know exactly how it got derived in dy and dB.

By the way, if necessary, I can provide some more informations of my book for contextualization purposes. Sorry if I'm sounding confused, I study economics by self-taught and this is my first time using Economics Stack Exchange.

Thanks for your attention!

**Tags:** macroeconomics, mathematical-economics

> **Comments:**
> **username_0**: Of course. I did not mention it before because it is not from any of the major economics textbooks and I thought that it would not be that helpful. It is "Inflação e Crises", by Affonso Celso Pastore.  
> **username_1**: What is it that you do not understand? Do you know what total differentation is?  
> **username_0**: I have a basic knowledge on total differentiation cause I'm studying economics by self-taught. I mean, I explicitly have 3 doubts about it that I thought that if someone explained properly this total differentiation, I could understand it. I will write in the comments these three and edit the post for more clarity. Sorry if Im sounding confused, it's been 7 days that I'm trying to understand it completely and also it's my first time using StackExchange.  
> **username_0**: First doubt: I would like to know if there is a particular reason that, in this total differentiation, we use the derivatives dy, dr, dB, dM and dG, but we do not use dT, since T is a variable that appears in the original equation.  Second doubt: although I can see how the "inner" derivation processes occur, I am not sure I can correctly interpret the signs (the "plusses" and the "minuses"). I mean, why C_a dM is a negative term while dG is a positive term? This doubt extend to the other variables as well.  
> **username_0**: The third one: I have a doubt in the derivation of the T variable in dy and dB. I don't know exactly how it got derived in dy and dB.  
> **username_2**: @username_0 I added that reference to the post. Whenever you mention some sources you should refer to them even if they are not widely known  

---

## 1 answers

### Answer by username_2 on 2021-04-22. Score: 0 | Accepted answer

Total differentiation means you are differentiating all variables in the expression. So if we start with

$$y = C\{[y + B - T(y + B)]; [M + \frac{B}{r}]\} + I(r) + G$$

we get:

$$ d y = C\_y^{'} dy + C\_y^{'} dB - C\_y^{'} dTdy - C\_y^{'} dTdB + C\_A^{'} dM + C\_A^{'} \left( \frac{dBr-Bdr}{r^2} \right) + I'dr + dG $$

Here you are using just basic calculus such as chain rule accordding to which $\frac{d}{dx}[y(u(x))] = \frac{dy}{du} \frac{du}{dy}$, plus you use the rule that with total differentiation $dy(x\_1,x\_2) = y\_1' dx\_1 +y\_2'dx\_2$ and also quotient rule $\frac{d}{dx}[F(x)/G(x)]= \frac{F'G(x) - F(x)G'}{G(x)^2}$.

Next you just move everything to the LHS of the equation to get

$$dy - C\_y^{'} dy - C\_y^{'} dB + C\_y^{'} T'dy + C\_y^{'} T'dB - C\_A^{'} dM - C\_A^{'} \left( \frac{dBr-Bdr}{r^2} \right) - I'dr - dG = 0$$

Now just collect terms and clean up the mess above and you directly get:

$$(1 + C\_y^{'}(1 - T'))dy - (I\_r - C\_A \frac{B}{r^2})dr - [C\_y (1 - T') + \frac{C\_A}{r}]dB- C\_A^{'} dM - dG = 0$$

The expression looks difficult because it has a lot of terms but you should not get intimidated by its look.

> **Comments:**
> **username_0**: That kind of clarification was exactly what I needed, thank you, muflon. By the way, I consulted my book and the "dG" term is written as a positive one, do you think it can be a mistake in the book? The way you developed the total differentiation here with the negative "dG" seems a lot more reasonable for me. Thanks again.  
> **username_2**: @username_0 you are welcome, if you think this answer answered your Q consider accepting it. I think that must have been a typo because except for moving it to other side we do not do any operation with dG so it should be negative  

---

