Intepretation of $dC_1=-\theta(dT_1+\frac{dT_2}{1+r})=0\Longleftrightarrow dT_1=-\frac{dT_2}{1+r}$? We are studying the Ricardian equivalence in a macro course and we are asked to show that a tax cut today will be matched by a tax hike tomorrow. I believe the above is supposed to be the solution, however I do not quite understand the total differential and even less so the notation. I have three questions:


How exactly do I take the total derivative of $C_1=\theta(V_1+Y^L_1-T_1+\frac{Y^L_2-T_2}{1+r})$ with respect to $T_1$ and $T_2$?
How would I then get to $dC_1=-\theta(dT_1+\frac{dT_2}{1+r})$? 
What is the interpretation of $dT_1$ and $dT_2$? Specifically, what is the interpretation of $dT_1=-\frac{dT_2}{1+r}$?
Basic intro to total differentiation

Suppose we have a function $$f(x_1,x_2,\ldots,x_n).$$ The total derivative is given by $$\frac{\partial f}{\partial x_1}dx_1+\frac{\partial f}{\partial x_2}dx_2+\ldots+\frac{\partial f}{\partial x_n}dx_n,$$
where the $\partial$ terms are partial derivatives and the $dx$ terms are so-called "differentials".

Conceptually, what is going on is the following: we ask "if I change every variable $x_i$ by a small amount $dx_i$, what is the effect on the overall value of $f$?"

To answer this question, we can think about the variables one-by-one. Start with $x_1$. A unit increase in $x_1$ causes $f$ to change by $\partial f/\partial x_1$. We are making a small $dx_1$ unit change in $x_1$. Thus, the total change in $f$ due to the change in $x_1$ is $\frac{\partial f}{\partial x_1}dx_1$. At the same time, we change $x_2$ by $dx_2$, resulting in a $\frac{\partial f}{\partial x_2}dx_2$ unit change in $f$ and so on.

If you want to read more on the topic, I'd recommend the relevant chapter of "Fundamental Methods of Mathematical Economics" by Alpha Chiang.



Aside on why this might be useful (can be skipped without loss)

Let's look at an example of how this could be useful. Suppose we have a game where players $i=1,2$ each choose an action $x_i$. Player $i$'s payoff is 

$$u_i=f(x_i,a)+g(x_i,x_j,a)$$

where $a$ is a parameter and $x_j$ is the other player's action. Thus, a player's best action is given by the first-order condition

$$f_1(x_i,a)+g_1(x_i,x_j,a)=0$$
where $f_i$ means the partial derivative of $f$ wrt its $i^{th}$ argument. 

It is hard to know how a change in the parameter $a$ will affect $i$'s choice of $x_i$ because the change in $a$ will affect $x_i$ not only directly, but also by changing $x_j$. When all of these changes have settled back to equilibrium, what happens? It seems like we can't go any further because we need to know the functional form of $f$ and $g$ to be able to explicitly solve for $x_i$ and see how they depend on each other and on $a$.

But total differentiation can help us understand the properties of the solution here. Let's totally differentiate the first order condition, allowing all of $x_i$, $x_j$, and $a$ to change at once (I omit the arguments of the functions for brevity):

$$[f_{11}+g_{11}]dx_i+g_{12}dx_j+[f_{12}+g_{13}]da=0.$$

Now I can solve for $\frac{dx_i}{da}$:

$$\frac{dx_i}{da}=-\frac{f_{12}+g_{13}}{f_{11}+g_{11}}.$$
So long as I know something about the individual properties of $f$ and $g$, I can use this equation to tell me what the sign of $fx_i/da$ will be!



Coming back to your example

For notational convenience, I give your function a name ($f$):

$$f(C_1,T_1,T_2)\equiv\theta(V_1+Y^L_1-T_1+\frac{Y^L_2-T_2}{1+r})-C_1=0,$$

we apply the basic rules of total differentiation:

$$f_1 dC_1+f_2dT_1+f_3dT_2=0.$$

We compute $f_1=\frac{\partial f}{\partial C_1}=-1$,$f_2=\frac{\partial f}{\partial T_1}=-\theta$, $f_3=\frac{\partial f}{\partial T_2}=-\theta\frac{1}{1+r}$. This therefore gives $$dC_1=-\theta\left(dT_1+\frac{dT_2}{1+r}\right).$$

In words, this equation tells us that the change in $C_1$ must be equal to $-\theta\left(dT_1+\frac{dT_2}{1+r}\right)$, where $dT_1$ and $dT_2$ are the changes in $T_1$ and $T_2$. Thus, if we want to say that $C_1$ doesn't change then it must be that $dT_1=-\frac{dT_2}{1+r}$ so that the changes in $T_1$ and $T_2$ 'cancel-out'.