# Total differentiation in a IS curve context

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.

• 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 Apr 22 at 18:59
• What is it that you do not understand? Do you know what total differentation is? – Giskard Apr 22 at 19:02
• 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 Apr 22 at 19:36
• 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 Apr 22 at 19:41
• 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 Apr 22 at 19:45

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.

• 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 Apr 22 at 23:07
• @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 Apr 22 at 23:12