<p>This is more of a calculus question. Recall the total differential of a function <span class="math-container">$f(z_1,z_2,z_3)$</span>:<br />
<span class="math-container">\begin{equation}
\mathrm df(z_1,z_2,z_3)=\frac{\partial f(z_1,z_2,z_3)}{\partial z_1}\mathrm dz_1 + \frac{\partial f(z_1,z_2,z_3)}{\partial z_2}\mathrm dz_2+\frac{\partial f(z_1,z_2,z_3)}{\partial z_3}\mathrm dz_3.
\end{equation}</span>
Hence,
<span class="math-container">\begin{equation}
\frac{\mathrm df(z_1,z_2,z_3)}{\mathrm dz_1}=\frac{\partial f(z_1,z_2,z_3)}{\partial z_1}\frac{\mathrm dz_1}{\mathrm dz_1} + \frac{\partial f(z_1,z_2,z_3)}{\partial z_2}\frac{\mathrm dz_2}{\mathrm dz_1} + \frac{\partial f(z_1,z_2,z_3)}{\partial z_3}\frac{\mathrm dz_3}{\mathrm dz_1}
\end{equation}</span>
Applying to your case, we substitute <span class="math-container">$p_1$</span> for <span class="math-container">$z_1$</span>, <span class="math-container">$p_2$</span> for <span class="math-container">$z_2$</span>, and <span class="math-container">$\overline m$</span> for <span class="math-container">$z_3$</span> (noting that <span class="math-container">$\overline m(p_1,p_2)$</span> is a function of the prices):<br />
<span class="math-container">\begin{align}\require{cancel}
\frac{\mathrm dx_1(p_1,p_2,\overline m)}{\mathrm dp_1}&amp;=\frac{\partial x_1(p_1,p_2,\overline m)}{\partial p_1}\cancelto{1}{\frac{\mathrm dp_1}{\mathrm dp_1}} + \frac{\partial x_1(p_1,p_2,\overline m)}{\partial p_2}\cancelto{0}{\frac{\mathrm dp_2}{\mathrm dp_1}} + \frac{\partial x_1(p_1,p_2,\overline m)}{\partial \overline m}\frac{\mathrm d\overline m}{\mathrm dp_1}\\
&amp;=\frac{\partial x_1(p_1,p_2,\overline m)}{\partial p_1} + \frac{\partial x_1(p_1,p_2,\overline m)}{\partial \overline m}\overline x_1
\end{align}</span></p>

CC BY-SA 4.0
2021-05-18T14:00:15.040Z
Sorry, I don't quite understand tbh. I think you mean "substitute $p_1$ for $z_1$"? Also, why does the total differential not include the partial derivative for $p_2$ but only for $p_1$ and $m$? And why is $p_1\overline{x}_1+p_2\overline{x}_2=\overline{m}$? Shouldn't this be m if $\overline{m}$ is the original budget and $p_1, p_2$ are the new prices? @Herr K.
j3141592653589793238
CC BY-SA 4.0
2021-05-18T17:07:21.300Z
Got it, nvm, it took a while but it finally made "pling" up there.
j3141592653589793238
CC BY-SA 4.0
2021-05-18T17:20:41.063Z
@j3141592653589793238: I edited in some intermediate steps. Hopefully things will be clearer. Regarding $\overline m$, I took that from the first identity in your post: $x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2) \equiv x_1(p_1,p_2,p_1\overline{x}_1+p_2\overline{x}_2)$, which defines the third argument of $x_1$ to be $p_1\overline{x}_1+p_2\overline{x}_2$ (whether you call this $\overline m$ or otherwise).
Herr K.
CC BY-SA 4.0
2021-05-18T12:41:22.547Z
2021-05-20T18:19:51.517Z
2021-05-20T18:19:51.517Z
Herr K.
<p>Why is the partial derivative of <span class="math-container">$x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2) \equiv x_1(p_1,p_2,p_1\overline{x}_1+p_2\overline{x}_2)$</span> for <span class="math-container">$p_1$</span>
<span class="math-container">$$
\frac{\partial x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2)}{\partial p_1}=\frac{\partial x_1(p_1,p_2,\overline{m})}{\partial p_1} +\frac{\partial x_1(p_1,p_2,\overline{m})}{\partial m}\overline{x}_1
$$</span></p>
<p>where <span class="math-container">$(\overline{x}_1,\overline{x}_2)$</span> is the bundle originally demanded at prices <span class="math-container">$(\overline{p}_1,\overline{p}_2)$</span> and income <span class="math-container">$\overline{m}$</span> and <span class="math-container">$x_1^S$</span> is the Slutsky demand function for good 1 and <span class="math-container">$x_1$</span> the Marshallian demand function for good 1?</p>
<p>I understand that <span class="math-container">$m=p_1\overline{x}_1+p_2\overline{x}_2$</span> because the pivot ensures that the consumer is still having enough income to purchase his old bundle <span class="math-container">$(\overline{x}_1,\overline{x}_2)$</span>.</p>
<p>Chapter 8 of Varian shows this to derive the Slutsky equation using calculus. I don't understand the right side of the equation.</p>
<p>I know that <span class="math-container">$p_1\overline{x}_1+p_2\overline{x}_2=m$</span> and <span class="math-container">$\overline{p}_1\overline{x}_1+\overline{p}_2\overline{x}_2=\overline{m}$</span> where <span class="math-container">$m$</span> is the new budget.</p>

<microeconomics><mathematical-economics><utility><slutsky-equation>
Derivative of $x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2) \equiv x_1(p_1,p_2,p_1\overline{x}_1 + p_2\overline{x}_2)$ to derive Slutsky equation
j3141592653589793238
# Derivative of $x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2) \equiv x_1(p_1,p_2,p_1\overline{x}_1 + p_2\overline{x}_2)$ to derive Slutsky equation

(Asked by: username_0 on 2021-05-17)

Why is the partial derivative of $x\_1^S(p\_1, p\_2, \overline{x}\_1, \overline{x}\_2) \equiv x\_1(p\_1,p\_2,p\_1\overline{x}\_1+p\_2\overline{x}\_2)$ for $p\_1$
$$
\frac{\partial x\_1^S(p\_1, p\_2, \overline{x}\_1, \overline{x}\_2)}{\partial p\_1}=\frac{\partial x\_1(p\_1,p\_2,\overline{m})}{\partial p\_1} +\frac{\partial x\_1(p\_1,p\_2,\overline{m})}{\partial m}\overline{x}\_1
$$

where $(\overline{x}\_1,\overline{x}\_2)$ is the bundle originally demanded at prices $(\overline{p}\_1,\overline{p}\_2)$ and income $\overline{m}$ and $x\_1^S$ is the Slutsky demand function for good 1 and $x\_1$ the Marshallian demand function for good 1?

I understand that $m=p\_1\overline{x}\_1+p\_2\overline{x}\_2$ because the pivot ensures that the consumer is still having enough income to purchase his old bundle $(\overline{x}\_1,\overline{x}\_2)$.

Chapter 8 of Varian shows this to derive the Slutsky equation using calculus. I don't understand the right side of the equation.

I know that $p\_1\overline{x}\_1+p\_2\overline{x}\_2=m$ and $\overline{p}\_1\overline{x}\_1+\overline{p}\_2\overline{x}\_2=\overline{m}$ where $m$ is the new budget.

**Tags:** microeconomics, mathematical-economics, utility, slutsky-equation

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## 1 answers

### Answer by username_1 on 2021-05-18. Score: 5 | Accepted answer

This is more of a calculus question. Recall the total differential of a function $f(z\_1,z\_2,z\_3)$:  
\begin{equation}
\mathrm df(z\_1,z\_2,z\_3)=\frac{\partial f(z\_1,z\_2,z\_3)}{\partial z\_1}\mathrm dz\_1 + \frac{\partial f(z\_1,z\_2,z\_3)}{\partial z\_2}\mathrm dz\_2+\frac{\partial f(z\_1,z\_2,z\_3)}{\partial z\_3}\mathrm dz\_3.
\end{equation}
Hence,
\begin{equation}
\frac{\mathrm df(z\_1,z\_2,z\_3)}{\mathrm dz\_1}=\frac{\partial f(z\_1,z\_2,z\_3)}{\partial z\_1}\frac{\mathrm dz\_1}{\mathrm dz\_1} + \frac{\partial f(z\_1,z\_2,z\_3)}{\partial z\_2}\frac{\mathrm dz\_2}{\mathrm dz\_1} + \frac{\partial f(z\_1,z\_2,z\_3)}{\partial z\_3}\frac{\mathrm dz\_3}{\mathrm dz\_1}
\end{equation}
Applying to your case, we substitute $p\_1$ for $z\_1$, $p\_2$ for $z\_2$, and $\overline m$ for $z\_3$ (noting that $\overline m(p\_1,p\_2)$ is a function of the prices):  
\begin{align}\require{cancel}
\frac{\mathrm dx\_1(p\_1,p\_2,\overline m)}{\mathrm dp\_1}&=\frac{\partial x\_1(p\_1,p\_2,\overline m)}{\partial p\_1}\cancelto{1}{\frac{\mathrm dp\_1}{\mathrm dp\_1}} + \frac{\partial x\_1(p\_1,p\_2,\overline m)}{\partial p\_2}\cancelto{0}{\frac{\mathrm dp\_2}{\mathrm dp\_1}} + \frac{\partial x\_1(p\_1,p\_2,\overline m)}{\partial \overline m}\frac{\mathrm d\overline m}{\mathrm dp\_1}\\
&=\frac{\partial x\_1(p\_1,p\_2,\overline m)}{\partial p\_1} + \frac{\partial x\_1(p\_1,p\_2,\overline m)}{\partial \overline m}\overline x\_1
\end{align}

> **Comments:**
> **username_0**: Sorry, I don't quite understand tbh. I think you mean "substitute $p_1$ for $z_1$"? Also, why does the total differential not include the partial derivative for $p_2$ but only for $p_1$ and $m$? And why is $p_1\overline{x}_1+p_2\overline{x}_2=\overline{m}$? Shouldn't this be m if $\overline{m}$ is the original budget and $p_1, p_2$ are the new prices? @Herr K.  
> **username_0**: Got it, nvm, it took a while but it finally made "pling" up there.  
> **username_1**: @j3141592653589793238: I edited in some intermediate steps. Hopefully things will be clearer. Regarding $\overline m$, I took that from the first identity in your post: $x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2) \equiv x_1(p_1,p_2,p_1\overline{x}_1+p_2\overline{x}_2)$, which defines the third argument of $x_1$ to be $p_1\overline{x}_1+p_2\overline{x}_2$ (whether you call this $\overline m$ or otherwise).  

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