Partial derivatives: Prove $fracdx_2dx_1=-fracMU_1MU_2$

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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 $displaystyle MU_1=fracpartial upartial x_1$ and $displaystyle MU_2=fracpartial upartial x_2$.

Suppose $u(x_1, x_2)=c$ for some constant $c$, show that $displaystylefracdx_2dx_1=-fracMU_1MU_2$.

partial-derivative

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edited Jul 31 at 20:37

asked Jul 31 at 19:53

Thomas

574313

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you worked out so far ?
  – Ahmad Bazzi
  Jul 31 at 19:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Dini's theorem?
  – edo1998
  Jul 31 at 20:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$
  – Rumpelstiltskin
  Jul 31 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adam Oops, I forgot about the $c$. I've edited my post.
  – Thomas
  Jul 31 at 20:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment
  – Rumpelstiltskin
  Jul 31 at 21:11
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

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 $displaystyle MU_1=fracpartial upartial x_1$ and $displaystyle MU_2=fracpartial upartial x_2$.

Suppose $u(x_1, x_2)=c$ for some constant $c$, show that $displaystylefracdx_2dx_1=-fracMU_1MU_2$.

partial-derivative

share|cite|improve this question

edited Jul 31 at 20:37

asked Jul 31 at 19:53

Thomas

574313

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you worked out so far ?
  – Ahmad Bazzi
  Jul 31 at 19:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Dini's theorem?
  – edo1998
  Jul 31 at 20:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$
  – Rumpelstiltskin
  Jul 31 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adam Oops, I forgot about the $c$. I've edited my post.
  – Thomas
  Jul 31 at 20:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment
  – Rumpelstiltskin
  Jul 31 at 21:11
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

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 $displaystyle MU_1=fracpartial upartial x_1$ and $displaystyle MU_2=fracpartial upartial x_2$.

Suppose $u(x_1, x_2)=c$ for some constant $c$, show that $displaystylefracdx_2dx_1=-fracMU_1MU_2$.

partial-derivative

share|cite|improve this question

edited Jul 31 at 20:37

asked Jul 31 at 19:53

Thomas

574313

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 $displaystyle MU_1=fracpartial upartial x_1$ and $displaystyle MU_2=fracpartial upartial x_2$.

Suppose $u(x_1, x_2)=c$ for some constant $c$, show that $displaystylefracdx_2dx_1=-fracMU_1MU_2$.

partial-derivative

share|cite|improve this question

edited Jul 31 at 20:37

asked Jul 31 at 19:53

Thomas

574313

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 20:37

edited Jul 31 at 20:37

edited Jul 31 at 20:37

asked Jul 31 at 19:53

Thomas

574313

asked Jul 31 at 19:53

Thomas

574313

asked Jul 31 at 19:53

Thomas

574313

574313

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you worked out so far ?
  – Ahmad Bazzi
  Jul 31 at 19:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Dini's theorem?
  – edo1998
  Jul 31 at 20:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$
  – Rumpelstiltskin
  Jul 31 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adam Oops, I forgot about the $c$. I've edited my post.
  – Thomas
  Jul 31 at 20:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment
  – Rumpelstiltskin
  Jul 31 at 21:11
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you worked out so far ?
  – Ahmad Bazzi
  Jul 31 at 19:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Dini's theorem?
  – edo1998
  Jul 31 at 20:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$
  – Rumpelstiltskin
  Jul 31 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adam Oops, I forgot about the $c$. I've edited my post.
  – Thomas
  Jul 31 at 20:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment
  – Rumpelstiltskin
  Jul 31 at 21:11
  
  

  
  
  
  

What have you worked out so far ?
– Ahmad Bazzi
Jul 31 at 19:58

What have you worked out so far ?
– Ahmad Bazzi
Jul 31 at 19:58

1

1

Do you know Dini's theorem?
– edo1998
Jul 31 at 20:05

Do you know Dini's theorem?
– edo1998
Jul 31 at 20:05

1

1

Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$
– Rumpelstiltskin
Jul 31 at 20:15

Is $x_2$ defined as solution to $u(x_1, x_2(x_1))=c$ for some constant $c$
– Rumpelstiltskin
Jul 31 at 20:15

@Adam Oops, I forgot about the $c$. I've edited my post.
– Thomas
Jul 31 at 20:35

@Adam Oops, I forgot about the $c$. I've edited my post.
– Thomas
Jul 31 at 20:35

1

1

It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment
– Rumpelstiltskin
Jul 31 at 21:11

It's called the implicit function theorem, which is also apparently called Dini's theorem in Italy, which explains edo1998's comment
– Rumpelstiltskin
Jul 31 at 21:11

add a comment | 

1 Answer
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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).

$displaystylefracdudx_1=fracpartial upartial x_1fracdx_1dx_1+fracpartial upartial x_2fracdx_2dx_1$

Since $u$ is a constant function of $x_1$ (only), $displaystylefracdudx_1=0$.

So, $displaystyle 0=MU_1cdot 1+MU_2fracdx_2dx_1$

$displaystyleimplies MU_2fracdx_2dx_1=-MU_1$

$displaystylefracdx_2dx_1=frac-MU_1MU_2$

QED

share|cite|improve this answer

edited Jul 31 at 22:17

answered Jul 31 at 21:01

Thomas

574313

add a comment | 

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1 Answer
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active

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votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
0
down vote

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).

$displaystylefracdudx_1=fracpartial upartial x_1fracdx_1dx_1+fracpartial upartial x_2fracdx_2dx_1$

Since $u$ is a constant function of $x_1$ (only), $displaystylefracdudx_1=0$.

So, $displaystyle 0=MU_1cdot 1+MU_2fracdx_2dx_1$

$displaystyleimplies MU_2fracdx_2dx_1=-MU_1$

$displaystylefracdx_2dx_1=frac-MU_1MU_2$

QED

share|cite|improve this answer

edited Jul 31 at 22:17

answered Jul 31 at 21:01

Thomas

574313

add a comment | 

up vote
0
down vote

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).

$displaystylefracdudx_1=fracpartial upartial x_1fracdx_1dx_1+fracpartial upartial x_2fracdx_2dx_1$

Since $u$ is a constant function of $x_1$ (only), $displaystylefracdudx_1=0$.

So, $displaystyle 0=MU_1cdot 1+MU_2fracdx_2dx_1$

$displaystyleimplies MU_2fracdx_2dx_1=-MU_1$

$displaystylefracdx_2dx_1=frac-MU_1MU_2$

QED

share|cite|improve this answer

edited Jul 31 at 22:17

answered Jul 31 at 21:01

Thomas

574313

add a comment | 

up vote
0
down vote

up vote
0
down vote

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).

$displaystylefracdudx_1=fracpartial upartial x_1fracdx_1dx_1+fracpartial upartial x_2fracdx_2dx_1$

Since $u$ is a constant function of $x_1$ (only), $displaystylefracdudx_1=0$.

So, $displaystyle 0=MU_1cdot 1+MU_2fracdx_2dx_1$

$displaystyleimplies MU_2fracdx_2dx_1=-MU_1$

$displaystylefracdx_2dx_1=frac-MU_1MU_2$

QED

share|cite|improve this answer

edited Jul 31 at 22:17

answered Jul 31 at 21:01

Thomas

574313

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).

$displaystylefracdudx_1=fracpartial upartial x_1fracdx_1dx_1+fracpartial upartial x_2fracdx_2dx_1$

Since $u$ is a constant function of $x_1$ (only), $displaystylefracdudx_1=0$.

So, $displaystyle 0=MU_1cdot 1+MU_2fracdx_2dx_1$

$displaystyleimplies MU_2fracdx_2dx_1=-MU_1$

$displaystylefracdx_2dx_1=frac-MU_1MU_2$

QED

share|cite|improve this answer

edited Jul 31 at 22:17

answered Jul 31 at 21:01

Thomas

574313

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 22:17

edited Jul 31 at 22:17

edited Jul 31 at 22:17

answered Jul 31 at 21:01

Thomas

574313

answered Jul 31 at 21:01

Thomas

574313

answered Jul 31 at 21:01

Thomas

574313

574313

add a comment | 

add a comment | 

 

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