# Is Marginal Rate of Substitution a multivariable function?

Suppose I have $U(x,y)$ and a level set of indifference curves. Suppose the value of $U$ along a given curve is $\bar{U}$. We know $dU = 0$. We compute total derivative, rearrange, and now have

$$\frac{dy}{dx} = -\frac{U_x}{U_y}$$

My Confusion and Question:

$\frac{dy}{dx}$ is written as a function of a single variable right? But when we talk about $MU$, we always talk about it as a multivariable function. I thought $MRS$ was the ordinal utility analog of the cardinal utility $MU$. So why is $MRS$ a single variable function?

## 2 Answers

$\frac{dy}{dx}$ is written as a function of a single variable right?

I don't think so. Your shorthand notation should be standing for

$$\frac{dy}{dx} = -\frac{U_x(x, y)}{U_y(x,y}$$

and hence is a function of both variables ${x,y}$. That makes sense, as your MRS could depend on the locus you're at.

• Okay. I buy what you are saying about the RHS being multivariable. But the LHS is clearly a single variable notation.....right? – Stan Shunpike Apr 10 '15 at 13:48
• @StanShunpike yes, the shorthand is abuse of notation. When introduced, one should be clear that. – FooBar Apr 10 '15 at 13:58
• But how can we abuse it like than and still consider them differentials? Like that just seems like it wouldn't make any sense. Is there a way to write it that doesn't involve this abuse of notation? – Stan Shunpike Apr 10 '15 at 14:00

We have the function

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

from which we create the equation

$$x^*,y^*: U(x^*,y^*) = \bar U$$

with $\bar U$ being fixed a constant, and so $d\bar U = 0$.

Then

$$dU(x^*,y^*) = d\bar U = 0 \implies U_x(x^*,y^*)dx^*+U_y(x^*,y^*)dy^* = 0$$

$$\implies \frac {dy^*}{dx^*} = -\frac {U_x(x^*,y^*)}{U_y(x^*,y^*)}$$

Leaving aside the (mathematical) "Civil War of differentials", there is no abuse of notation here, only simplification for reasons of compactness. Why? $x^*$ and $y^*$ are solutions to the equation, not variables free to vary independently.So in reality we have $y^* = h(x^*,\bar U)$ and

$$\implies \frac {dh(x^*,\bar U)}{dx^*} = -\frac {U_x(x^*,h(x^*,\bar U))}{U_y(x^*,h(x^*,\bar U))}$$

NOTE: these kinds of notation simplifications appear in all disciplines that use mathematics. They are not always beneficial though: for example M. Caputo in Foundations of Dynamic Economic Analysis more than most writes explicitly all dependencies. This makes notation cumbersome and confusing at first, but once you get the hang of it, you are never lost.