<p>This is intended to be a partial answer. I hope more knowledgeable people will answer. </p>

<p>Apart from asserting that $\exists \phi(\mathbf{x}) =\mathbf{y}$ the implicit function theorem also asserts </p>

<p>\begin{equation}
D \phi(\mathbf{x}) = - \left(\frac{\partial U}{\partial \mathbf{y}} \left( \begin{array}{c} x\\ \phi(\mathbf{x}) \\\end{array} \right) \right)^{-1}\left( \frac{\partial U}{\partial \mathbf{x}} \left( \begin{array}{c} x\\ \phi(\mathbf{x}) \\\end{array} \right) \right)
\end{equation}
in this case that corresponds to </p>

<p>$$\frac{d \phi(x)}{dx} = - \left(\frac{1}{y}\right)^{-1}\left(\frac{1}{x}\right) = -\frac{y}{x}$$
so thus 
$$ \frac{dy}{dx}= -\frac{y}{x}$$</p>

<p>This shows that the definition of $MRS$ follows from the Implicit Function Theorem. </p>

<p>EDIT 1: </p>

<p>Consider $$ dy = - (U_y)^{-1} U_x dx$$ which we agree follows from the implicit function theorem. Then by multiplying both sides by $U_y$ and rearranging we have </p>

<p>\begin{align*}
U_y dy &amp;=  -U_x dx \\
U_x dx + U_y dy &amp;= 0 
\end{align*}
and thus by the definition of a differential, </p>

<p>$$ U_x dx + U_y dy = 0 = dU$$</p>

<p>so this is also a consequence of the implicit function theorem. This can also be seen trivially since $U = 0$, then by the definition of a differential, $dU =0$. </p>

CC BY-SA 3.0
2015-05-10T18:40:35.027Z
2015-05-10T23:10:19.147Z
2015-05-10T23:10:19.147Z
Stan Shunpike
<p>It's actually pretty straight forward. The implicit function theorem for two variables is given as follows (as long as some regularity conditions hold):</p>

<p>For $F(x, y) = 0$,</p>

<p>$
\frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}
$</p>

<p>In the case of <em>MRS</em>, we want the marginal change in $x$ associated with a marginal change in $y$ required to maintain a certain level of utility, $c$, such as (conveniently) $c=0$. So, starting with</p>

<p>$U = U(x, y) = 0$,</p>

<p>we have</p>

<p>$\frac{dy}{dx} = -\frac{U_x}{U_y}$</p>

<p>Note that $c=0$ is just a simplification for exposition. For a general $c$, you can just subtract it from either side of the equation and you get the same result since $c$ disappears in the derivative.</p>

CC BY-SA 3.0
2016-02-16T15:19:48.693Z
And utility satisfies such regularity conditions?
BCLC
CC BY-SA 3.0
2016-02-16T15:30:06.320Z
@BCLC Usually. The two main conditions are 1) $U(.,.)$ needs to be continuously differentiable and 2) the partial of F wrt y needs to be nonzero. The second is generally innocuous. The first is usually assumed either as a general primitive or by functional form, with the primary exception being behavioral econ. studies that break the assumption intentionally (eg. Prospect Theory)
philE
CC BY-SA 3.0
2015-05-13T03:18:31.013Z
2015-05-13T17:06:35.560Z
2015-05-13T17:06:35.560Z
philE
<p>TRAIN OF THOUGHT 1: </p>

<p>From what I understand, $MRS$ is calculated as </p>

<p>$$dU = U_x dx +  U_y dy =0$$ 
which by rearrangement yields 
$$\frac{dy}{dx}= -\frac{U_x}{U_y}$$</p>

<p>So suppose I have $$U(x,y) = \ln x +\ln y$$ Then 
$$ \frac{dy}{dx}= -\frac{1/x}{1/y} = -\frac{y}{x}$$</p>

<p>Okay. So I have a function $y$ in terms of $x$.</p>

<p>TRAIN OF THOUGHT 2: </p>

<p>Now consider my $U(x,y)$ again. Let $$\mathbf{a} = \begin{bmatrix}
1\\
1
\end{bmatrix}$$ 
and $$U(\mathbf{a})=0$$
We have $$DU(x,y) = \begin{bmatrix}
\frac{1}{x} &amp; \frac{1}{y}
\end{bmatrix} $$
and $$\frac{\partial U}{\partial y} (\mathbf{a}) = \begin{bmatrix}\frac{1}{y}\end{bmatrix}= 1$$ which is nonsingular since $\det(1) = 1$ and so by the Implicit Function Theorem, $$U = 0$$ defines $y$ implicitly as a function of $x$ in a neighborhood of $\mathbf{a}$.  </p>

<p><strong>My Question:</strong></p>

<p>How are these two trains of thought connected? The first is stated in terms of differentials. But the second is not. So I am confused why the definition of $MRS$ follows from the implicit function theorem. </p>

<consumer-theory>
Why does the definition of MRS follow from the implicit function theorem?
Stan Shunpike
# Why does the definition of MRS follow from the implicit function theorem?

(Asked by: username_0 on 2015-05-10)

TRAIN OF THOUGHT 1:

From what I understand, $MRS$ is calculated as

$$dU = U\_x dx + U\_y dy =0$$
which by rearrangement yields
$$\frac{dy}{dx}= -\frac{U\_x}{U\_y}$$

So suppose I have $$U(x,y) = \ln x +\ln y$$ Then
$$ \frac{dy}{dx}= -\frac{1/x}{1/y} = -\frac{y}{x}$$

Okay. So I have a function $y$ in terms of $x$.

TRAIN OF THOUGHT 2:

Now consider my $U(x,y)$ again. Let $$\mathbf{a} = \begin{bmatrix}
1\\
1
\end{bmatrix}$$
and $$U(\mathbf{a})=0$$
We have $$DU(x,y) = \begin{bmatrix}
\frac{1}{x} & \frac{1}{y}
\end{bmatrix} $$
and $$\frac{\partial U}{\partial y} (\mathbf{a}) = \begin{bmatrix}\frac{1}{y}\end{bmatrix}= 1$$ which is nonsingular since $\det(1) = 1$ and so by the Implicit Function Theorem, $$U = 0$$ defines $y$ implicitly as a function of $x$ in a neighborhood of $\mathbf{a}$.

**My Question:**

How are these two trains of thought connected? The first is stated in terms of differentials. But the second is not. So I am confused why the definition of $MRS$ follows from the implicit function theorem.

**Tags:** consumer-theory

---

## 2 answers

### Answer by username_1 on 2015-05-13. Score: 3 | Accepted answer

It's actually pretty straight forward. The implicit function theorem for two variables is given as follows (as long as some regularity conditions hold):

For $F(x, y) = 0$,

$
\frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}
$

In the case of *MRS*, we want the marginal change in $x$ associated with a marginal change in $y$ required to maintain a certain level of utility, $c$, such as (conveniently) $c=0$. So, starting with

$U = U(x, y) = 0$,

we have

$\frac{dy}{dx} = -\frac{U\_x}{U\_y}$

Note that $c=0$ is just a simplification for exposition. For a general $c$, you can just subtract it from either side of the equation and you get the same result since $c$ disappears in the derivative.

> **Comments:**
> **username_2**: And utility satisfies such regularity conditions?  
> **username_1**: @BCLC Usually. The two main conditions are 1) $U(.,.)$ needs to be continuously differentiable and 2) the partial of F wrt y needs to be nonzero. The second is generally innocuous. The first is usually assumed either as a general primitive or by functional form, with the primary exception being behavioral econ. studies that break the assumption intentionally (eg. Prospect Theory)  

---

### Answer by username_0 on 2015-05-10. Score: 2

This is intended to be a partial answer. I hope more knowledgeable people will answer.

Apart from asserting that $\exists \phi(\mathbf{x}) =\mathbf{y}$ the implicit function theorem also asserts

\begin{equation}
D \phi(\mathbf{x}) = - \left(\frac{\partial U}{\partial \mathbf{y}} \left( \begin{array}{c} x\\ \phi(\mathbf{x}) \\\end{array} \right) \right)^{-1}\left( \frac{\partial U}{\partial \mathbf{x}} \left( \begin{array}{c} x\\ \phi(\mathbf{x}) \\\end{array} \right) \right)
\end{equation}
in this case that corresponds to

$$\frac{d \phi(x)}{dx} = - \left(\frac{1}{y}\right)^{-1}\left(\frac{1}{x}\right) = -\frac{y}{x}$$
so thus
$$ \frac{dy}{dx}= -\frac{y}{x}$$

This shows that the definition of $MRS$ follows from the Implicit Function Theorem.

EDIT 1:

Consider $$ dy = - (U\_y)^{-1} U\_x dx$$ which we agree follows from the implicit function theorem. Then by multiplying both sides by $U\_y$ and rearranging we have

\begin{align\*}
U\_y dy &= -U\_x dx \\
U\_x dx + U\_y dy &= 0
\end{align\*}
and thus by the definition of a differential,

$$ U\_x dx + U\_y dy = 0 = dU$$

so this is also a consequence of the implicit function theorem. This can also be seen trivially since $U = 0$, then by the definition of a differential, $dU =0$.


---

