<p><strong>Basic intro to total differentiation</strong></p>

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

<p>Conceptually, what is going on is the following: we ask "if I change every variable <span class="math-container">$x_i$</span> by a <em>small</em> amount <span class="math-container">$dx_i$</span>, what is the effect on the overall value of <span class="math-container">$f$</span>?"</p>

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

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

<hr>

<p><strong>Aside on why this might be useful (can be skipped without loss)</strong></p>

<p>Let's look at an example of how this could be useful. Suppose we have a game where players <span class="math-container">$i=1,2$</span> each choose an action <span class="math-container">$x_i$</span>. Player <span class="math-container">$i$</span>'s payoff is </p>

<p><span class="math-container">$$u_i=f(x_i,a)+g(x_i,x_j,a)$$</span></p>

<p>where <span class="math-container">$a$</span> is a parameter and <span class="math-container">$x_j$</span> is the other player's action. Thus, a player's best action is given by the first-order condition</p>

<p><span class="math-container">$$f_1(x_i,a)+g_1(x_i,x_j,a)=0$$</span>
where <span class="math-container">$f_i$</span> means the partial derivative of <span class="math-container">$f$</span> wrt its <span class="math-container">$i^{th}$</span> argument. </p>

<p>It is hard to know how a change in the parameter <span class="math-container">$a$</span> will affect <span class="math-container">$i$</span>'s choice of <span class="math-container">$x_i$</span> because the change in <span class="math-container">$a$</span> will affect <span class="math-container">$x_i$</span> not only directly, but also by changing <span class="math-container">$x_j$</span>. 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 <span class="math-container">$f$</span> and <span class="math-container">$g$</span> to be able to explicitly solve for <span class="math-container">$x_i$</span> and see how they depend on each other and on <span class="math-container">$a$</span>.</p>

<p>But total differentiation can help us understand the properties of the solution here. Let's totally differentiate the first order condition, allowing all of <span class="math-container">$x_i$</span>, <span class="math-container">$x_j$</span>, and <span class="math-container">$a$</span> to change at once (I omit the arguments of the functions for brevity):</p>

<p><span class="math-container">$$[f_{11}+g_{11}]dx_i+g_{12}dx_j+[f_{12}+g_{13}]da=0.$$</span></p>

<p>Now I can solve for <span class="math-container">$\frac{dx_i}{da}$</span>:</p>

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

<hr>

<p><strong>Coming back to your example</strong></p>

<p>For notational convenience, I give your function a name (<span class="math-container">$f$</span>):</p>

<p><span class="math-container">$$f(C_1,T_1,T_2)\equiv\theta(V_1+Y^L_1-T_1+\frac{Y^L_2-T_2}{1+r})-C_1=0,$$</span></p>

<p>we apply the basic rules of total differentiation:</p>

<p><span class="math-container">$$f_1 dC_1+f_2dT_1+f_3dT_2=0.$$</span></p>

<p>We compute <span class="math-container">$f_1=\frac{\partial f}{\partial C_1}=-1$</span>,<span class="math-container">$f_2=\frac{\partial f}{\partial T_1}=-\theta$</span>, <span class="math-container">$f_3=\frac{\partial f}{\partial T_2}=-\theta\frac{1}{1+r}$</span>. This therefore gives <span class="math-container">$$dC_1=-\theta\left(dT_1+\frac{dT_2}{1+r}\right).$$</span></p>

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

CC BY-SA 4.0
2018-11-29T09:01:26.843Z
2018-11-29T10:51:48.953Z
2018-11-29T10:51:48.953Z
Ubiquitous
<p>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:</p>

<ol>
<li><p>How exactly do I take the total derivative of <span class="math-container">$C_1=\theta(V_1+Y^L_1-T_1+\frac{Y^L_2-T_2}{1+r})$</span> with respect to <span class="math-container">$T_1$</span> and <span class="math-container">$T_2$</span>?</p></li>
<li><p>How would I then get to <span class="math-container">$dC_1=-\theta(dT_1+\frac{dT_2}{1+r})$</span>? </p></li>
<li><p>What is the interpretation of <span class="math-container">$dT_1$</span> and <span class="math-container">$dT_2$</span>? Specifically, what is the interpretation of <span class="math-container">$dT_1=-\frac{dT_2}{1+r}$</span>?</p></li>
</ol>

CC BY-SA 4.0
2018-11-29T03:52:38.107Z
What's $V_1$ and $Y_1^L$ , as in what's their interpretation? Are these constants or they can change if tax changes? What I can interpret is they are fixed or given.
Henam
CC BY-SA 4.0
2018-11-29T07:44:31.780Z
$V_1$ is initial wealth in period 1 and $Y^L_1$ is income in period 1. They do not change with $T$.
Chisq
<macroeconomics>
Intepretation of $dC_1=-\theta(dT_1+\frac{dT_2}{1+r})=0\Longleftrightarrow dT_1=-\frac{dT_2}{1+r}$?
Chisq
# Intepretation of $dC_1=-\theta(dT_1+\frac{dT_2}{1+r})=0\Longleftrightarrow dT_1=-\frac{dT_2}{1+r}$?

(Asked by: username_0 on 2018-11-28)

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:

1. 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 $T\_2$?
2. How would I then get to $dC\_1=-\theta(dT\_1+\frac{dT\_2}{1+r})$?
3. What is the interpretation of $dT\_1$ and $dT\_2$? Specifically, what is the interpretation of $dT\_1=-\frac{dT\_2}{1+r}$?

**Tags:** macroeconomics
> **Comments:**
> **username_1**: What's $V_1$ and $Y_1^L$ , as in what's their interpretation? Are these constants or they can change if tax changes? What I can interpret is they are fixed or given.  
> **username_0**: $V_1$ is initial wealth in period 1 and $Y^L_1$ is income in period 1. They do not change with $T$.  

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## 1 answers

### Answer by username_2 on 2018-11-29. Score: 3 | Accepted answer

**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'.


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