Partial derivatives: Prove $\frac{dx_2}{dx_1}=-\frac{MU_1}{MU_2}$

Question

This is actually an economics question but it involves partial derivatives, so I thought it would be better to ask it here.

Let $u(x_1, x_2)$ be a function of 2 variables.

Let $\displaystyle MU_1=\frac{\partial u}{\partial x_1}$ and $\displaystyle MU_2=\frac{\partial u}{\partial x_2}$.

Suppose $u(x_1, x_2)=c$ for some constant $c$, show that $\displaystyle\frac{dx_2}{dx_1}=-\frac{MU_1}{MU_2}$.

Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$ — Jakobian, Jul 31 '18 at 20:15
It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment — Jakobian, Jul 31 '18 at 21:11

Siddhartha · Answer 1 · 2018-07-31T22:17:03.647

I think I've got it!

Consider the equation $u(x_1, x_2)=c$. Let $x_2=f(x_1)$.

Then $u$ is a differentiable function of $x_1$ (only).

$\displaystyle\frac{du}{dx_1}=\frac{\partial u}{\partial x_1}\frac{dx_1}{dx_1}+\frac{\partial u}{\partial x_2}\frac{dx_2}{dx_1}$

Since $u$ is a constant function of $x_1$ (only), $\displaystyle\frac{du}{dx_1}=0$.

So, $\displaystyle 0=MU_1\cdot 1+MU_2\frac{dx_2}{dx_1}$

$\displaystyle\implies MU_2\frac{dx_2}{dx_1}=-MU_1$

$\displaystyle\frac{dx_2}{dx_1}=\frac{-MU_1}{MU_2}$

QED

1 Answers1