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Implicit Differentiation

The level set $f(x,y) = z$ defines a “contour line” that can be plotted in $x-y$ space. In this section we derive the slope of such a contour line, $dy/dx$. This will turn out to have important economic implications in a wide range of applications.

The total derivative along a path

Before we think about level sets, let’s think about the more general question of how altitude varies along a path over the surface of a multivariable function. For example, let’s consider the path defined by $\textcolor{#31a354}{y(x) = 4 - 0.4x}$. We can draw that as a green line over the surface of a function, as shown below.

Now think about moving by some amount $\Delta x$ along the path, from some point $\textcolor{#3182bd}{(x,y)}$ to a second point $\textcolor{#d62728}{(x + \Delta x, y + \Delta y)}$. We can decompose this overall change into two changes:

  1. holding $y$ constant, from $\textcolor{#3182bd}{(x,y)}$ to $\textcolor{#e6550d}{(x+\Delta x, y)}$
  2. holding $x$ constant, from $\textcolor{#e6550d}{(x+\Delta x, y)}$ to $\textcolor{#d62728}{(x + \Delta x, y + \Delta y)}$

If we think about this in terms of partial derivatives, we can approximate the change due to $\Delta x$ as the change in $f(x,y)$ per unit change in $x$ (i.e., the partial derivative with respect to $x$), multiplied by $\Delta x$:

$$\left.\Delta f(x,y)\right|_{\Delta x} \approx {\partial f \over \partial x} \times \Delta x$$

Likewise, the change due to $\Delta y$ as

$$\left.\Delta f(x,y)\right|_{\Delta y} \approx {\partial f \over \partial y} \times \Delta y$$

Therefore the total change is the sum of these two changes:

$$\Delta f(x,y) \approx {\partial f \over \partial x} \times \Delta x + {\partial f \over \partial y} \times \Delta y$$

If we divide both sides by $\Delta x$, this becomes

$${\Delta f(x,y) \over \Delta x} \approx {\partial f \over \partial x} + {\partial f \over \partial y} \times {\Delta y \over \Delta x}$$

As $\Delta x \rightarrow 0$ in the limit, $\Delta y/\Delta x$ approaches the derivative of $y$ with respect to $x$, giving us

$$\left.{\partial f \over \partial x}\right|_{y = y(x)} = {\partial f \over \partial x} + {\partial f \over \partial y} \times {dy \over dx}$$

This is really just the chain rule: if $y$ changes when $x$ changes, then the total change in $f$ when $x$ changes is the direct effect due to the change in $x$, plus the indirect effect of the change in $y$.

The slope along a level set

The above analysis holds for any path along the surface. We can look in particular at the same analysis for a level set, which is implicitly defined by the equation \(f(x,y) = z\) where $z$ is some constant.

Taking the derivative of both sides of this equation with respect to $x$ gives us \({\partial f \over \partial x} + {\partial f \over \partial y} \times {dy \over dx} = 0\) The left-hand side of the equation comes from the analysis above; the right-hand side is zero because $z$ is a constant. Intuitively, along a level set, we know that the total change is zero: however much $f(x,y)$ increases as a result of $\Delta x$, it decreases by the same amount as a result of $\Delta y$:

As $\Delta x \rightarrow 0$, the blue line becomes a line tangent to the function at the point $(x, y, z)$. This allows us to solve for the slope along a level set at a point, by solving for $dy/dx$: \(\left.{dy \over dx}\right|_{f(x,y) = z} = - {\partial f/\partial x \over \partial f/\partial y}\) We can see if we plot level sets and contour maps, that we can use this formula to define the slope of the level set passing through any point $(x, y, f(x,y))$:

Importantly, note that every point $(x,y)$ defines a level set, and therefore the slope of a level set. No matter where you drag the blue point in this diagram, it defines a purple curve, and there is a line tangent to that curve at that point. Put another way, we can think of the level set itself and the slope of the level set as functions of the point $(x,y)$: \(\begin{aligned} \text{Level set through }(\hat x, \hat y) &= \{(x,y) | f(x,y) = f(\hat x, \hat y)\}\\ \text{Slope of that level set at }(\hat x, \hat y) &= \left.{dy \over dx}\right|_{f(x,y) = f(\hat x, \hat y)} = -{\partial f(\hat x, \hat y)/\partial x \over \partial f(\hat x, \hat y)/\partial y} \end{aligned}\)

These expressions will be incredibly important as we evaluate choices economic agents make, so be sure you are fluent in their applications.

Copyright (c) Christopher Makler /