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 $x$.


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.


2 Answers 2


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.

  • $\begingroup$ And utility satisfies such regularity conditions? $\endgroup$
    – BCLC
    Commented Feb 16, 2016 at 15:19
  • 1
    $\begingroup$ @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) $\endgroup$
    – philE
    Commented Feb 16, 2016 at 15:30

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.


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, $dU =0$.


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