Total differentiation in a IS curve context

Question

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!

Answer 1

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.