# 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

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.

This is more of a calculus question. Recall the total differential of a function $$f(z_1,z_2,z_3)$$:
$$$$\mathrm df(z_1,z_2,z_3)=\frac{\partial f(z_1,z_2,z_3)}{\partial z_1}\mathrm dz_1 + \frac{\partial f(z_1,z_2,z_3)}{\partial z_2}\mathrm dz_2+\frac{\partial f(z_1,z_2,z_3)}{\partial z_3}\mathrm dz_3.$$$$ Hence, $$$$\frac{\mathrm df(z_1,z_2,z_3)}{\mathrm dz_1}=\frac{\partial f(z_1,z_2,z_3)}{\partial z_1}\frac{\mathrm dz_1}{\mathrm dz_1} + \frac{\partial f(z_1,z_2,z_3)}{\partial z_2}\frac{\mathrm dz_2}{\mathrm dz_1} + \frac{\partial f(z_1,z_2,z_3)}{\partial z_3}\frac{\mathrm dz_3}{\mathrm dz_1}$$$$ Applying to your case, we substitute $$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}
• 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. May 18, 2021 at 14:00
• @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). May 18, 2021 at 17:20