Lecture 5: The Implicit Function Theorem

KSM3E04: Analysis on Manifolds — Fall 2025   |   August 22, 12:00–1:00 PM

In this lecture, we explore a fundamental tool in multivariable calculus: the Implicit Function Theorem. The theorem provides precise conditions under which a system of equations can be locally solved for certain variables in terms of others, allowing us to "untangle" implicit relationships. We begin with motivating examples that illustrate when and why such local solutions exist, then formally state and prove the theorem, and conclude by showing how to compute derivatives of functions defined implicitly.

1. Introduction: Explicit vs. Implicit Descriptions

Many equations in mathematics relate several variables, and a natural question arises: can one variable be expressed as a function of the others? This lecture explores when such an explicit representation is possible, even when a relationship is initially given in implicit form.

Consider the classical example of a circle of radius 1:

$$ f(x,y) = x^2 + y^2 - 1 = 0. $$

This equation implicitly defines a relationship between $x$ and $y$. At a point $(a,b)$ on the circle with $b>0$, we can locally write $y$ as an explicit function of $x$, namely $g(x) = \sqrt{1-x^2}$. For $b<0$, we use $g(x) = -\sqrt{1-x^2}$. However, this approach breaks down at $(\pm 1,0)$, where the tangent is vertical and a single $y$-value cannot be assigned to each nearby $x$.

This simple example illustrates the power of the Implicit Function Theorem: it tells us precisely when such a local explicit description exists. More generally, the theorem allows us to "untangle" implicit relationships and represent complex geometric objects, including boundaries, as graphs of functions in a neighborhood.

Recall from Lecture 3 the notion of a local defining function $r$ for an open set $\Omega \subset \mathbf{R}^n$ near a boundary point $p \in \partial \Omega$. For ready reference, a function $r: U \to \mathbf{R}$ defined on a neighborhood $U$ of $p$ is a local defining function if $$ \Omega \cap U = \{\mathbf{x} \in U \mid r(\mathbf{x}) < 0 \}, \qquad \partial \Omega \cap U = \{\mathbf{x} \in U \mid r(\mathbf{x}) = 0 \}, $$ and $$ dr(\mathbf{x}) \neq 0 \quad \text{for all } \mathbf{x} \in \partial \Omega \cap U. $$

The condition $dr \neq 0$ ensures that the boundary $\partial \Omega$ is captured smoothly and regularly by the level set $\{r=0\}$, inheriting a manifold structure of the same differentiability class as $r$. In other words, $dr \neq 0$ allows us to describe the boundary implicitly and guarantees that it can be locally represented as the graph of a function.

Thus, the Implicit Function Theorem and the concept of a local defining function together provide a practical and powerful framework for describing and analyzing boundaries and constraint sets in a rigorous way.

2. The Implicit Function Theorem

Consider a continuously differentiable function $f: \mathbf{R}^n \times \mathbf{R}^m \to \mathbf{R}^m$, with variables split as $n$ parameters $\mathbf{x}$ and $m$ dependent variables $\mathbf{y}$. Given a solution $(\mathbf{a}, \mathbf{b})$ of $f(\mathbf{x}, \mathbf{y}) = 0$, the natural question is whether we can locally solve for $\mathbf{y}$ as a function of $\mathbf{x}$ near $\mathbf{a}$. That is, does there exist a differentiable function $g$ with $\mathbf{y} = g(\mathbf{x})$ satisfying

$$ f(\mathbf{x}, g(\mathbf{x})) = 0 \,. $$

Theorem: Implicit Function Theorem

Let $f: \mathbf{R}^n \times \mathbf{R}^m \to \mathbf{R}^m$ be continuously differentiable near $(\mathbf{a}, \mathbf{b})$ with $f(\mathbf{a}, \mathbf{b}) = 0$. Let

$$ M = \frac{\partial f}{\partial \mathbf{y}}(\mathbf{a}, \mathbf{b}) = \big(D_{n+j} f^i(\mathbf{a}, \mathbf{b})\big)_{1 \le i,j \le m}. $$

If $\det M \neq 0$, then there exist open sets $A \subset \mathbf{R}^n$ containing $\mathbf{a}$ and $B \subset \mathbf{R}^m$ containing $\mathbf{b}$, and a unique differentiable function $g: A \to B$ such that

$$ f(\mathbf{x}, g(\mathbf{x})) = 0 \quad \text{for all } \mathbf{x} \in A. $$
Click to expand the full proof

Define the function $F: \mathbf{R}^n \times \mathbf{R}^m \to \mathbf{R}^n \times \mathbf{R}^m$ by

$$ F(\mathbf{x}, \mathbf{y}) = (\mathbf{x}, f(\mathbf{x}, \mathbf{y})). $$

Its Jacobian at $(\mathbf{a}, \mathbf{b})$ has block form

$$ F'(\mathbf{a}, \mathbf{b}) = \begin{pmatrix} I_n & 0 \\ * & M \end{pmatrix}. $$

Since $\det M \neq 0$, $F'(\mathbf{a}, \mathbf{b})$ is invertible. By the Inverse Function Theorem, there exist neighborhoods $U \subset \mathbf{R}^n$ and $V \subset \mathbf{R}^m$ such that $F$ has a differentiable inverse $$ h: F(U \times V) \to U \times V. $$

Observe that by the definition of $F$, its inverse $h$ must preserve the first $n$ coordinates. That is, $h$ can be written in the form

$$ h(\mathbf{x}, \mathbf{y}) = (\mathbf{x}, k(\mathbf{x}, \mathbf{y})) $$

for some differentiable function $k$. To solve $f(\mathbf{x}, \mathbf{y}) = 0$, we set $F(\mathbf{x}, \mathbf{y}) = (\mathbf{x}, 0)$, which gives

$$(\mathbf{x}, g(\mathbf{x})) = h(\mathbf{x}, 0) = (\mathbf{x}, k(\mathbf{x}, 0)).$$

Hence, the desired implicit function is

$$ g(\mathbf{x}) := k(\mathbf{x}, 0). $$

By construction, $g$ is differentiable and satisfies $f(\mathbf{x}, g(\mathbf{x})) = 0$. Uniqueness follows from the uniqueness of $h$.

Derivative of $g$: Differentiating $f(\mathbf{x}, g(\mathbf{x})) = 0$ with respect to $x^j$ gives

$$ \frac{\partial f}{\partial x^j} + M \cdot \frac{\partial g}{\partial x^j} = 0, $$ so that $$ g'(\mathbf{x}) = -\Big[\frac{\partial f}{\partial \mathbf{y}}\Big]^{-1} \Big[\frac{\partial f}{\partial \mathbf{x}}\Big]. $$

Example:

For $f(x,y) = x^2 + y^2 - 1$, we have $D_y f = 2y \neq 0$ wherever $y \neq 0$. Then

$$ g'(x) = -\frac{x}{g(x)}. $$

3. Topological and Smooth Manifolds

Building on our study of implicitly defined boundaries, we now introduce the concepts of topological and smooth manifolds, which generalize the idea of spaces that locally resemble Euclidean space.

Definition: Topological Manifold

An n-dimensional topological manifold is a Hausdorff, second-countable topological space $M$ in which every point $p \in M$ has a neighborhood $U$ homeomorphic to an open subset of $\mathbf{R}^n$. Such a homeomorphism $\varphi: U \to \varphi(U) \subset \mathbf{R}^n$ is called a chart, and a collection of charts that covers $M$ is called an atlas. Charts provide local coordinates, and the atlas organizes these charts to describe the manifold globally.

Definition: Smooth Manifold

A smooth n-dimensional manifold is a topological manifold $M$ together with an atlas of charts $\{(U_\alpha, \varphi_\alpha)\}$ such that whenever two charts overlap, the corresponding transition map $$ \varphi_\beta \circ \varphi_\alpha^{-1}: \varphi_\alpha(U_\alpha \cap U_\beta) \to \varphi_\beta(U_\alpha \cap U_\beta) $$ is a smooth (infinitely differentiable) function between open subsets of $\mathbf{R}^n$. In other words, the coordinates provided by different charts are compatible in a smooth way.

A concrete example arises from smoothly defined domains in $\mathbf{R}^n$. Let $\Omega \subset \mathbf{R}^n$ be open, and suppose its boundary $\partial \Omega$ is the zero set of a smooth function $r: \mathbf{R}^n \to \mathbf{R}$ with $dr \neq 0$ on $\partial \Omega$. By the Implicit Function Theorem, near each point $p \in \partial \Omega$, we can locally solve $r(x) = 0$ for one variable in terms of the others, producing a smooth parametrization of the boundary. These parametrizations serve as charts that give $\partial \Omega$ the structure of a smooth $(n-1)$-dimensional manifold.

Example: The unit sphere in $\mathbf{R}^3$, defined by $x^2 + y^2 + z^2 = 1$, is a 2-dimensional smooth manifold. Near any point where $z \neq \pm 1$, we can locally solve for $z$ as a function of $x$ and $y$, producing a smooth chart of the surface.

These ideas show how the boundaries of smoothly defined domains in $\mathbf{R}^n$ naturally carry a manifold structure. We will study smooth manifolds in much greater depth in later lectures, exploring their geometry, tangent spaces, and other fundamental concepts.

4. Problems for Practice

Exercise 1: Exploring a Non-Trivial Implicit Function

Consider the implicit equation $$ x e^y + y e^x = 0. $$

  1. Existence vs. Explicit Solutions: Explain why we cannot generally write an explicit solution of the form $y=g(x)$ for this equation. Then, use the Implicit Function Theorem to rigorously prove that a smooth, single-valued function $g(x)$ satisfying $f(x, g(x))=0$ exists in an open neighborhood of the point $(0,0)$. This is a core reason the IFT is so powerful—it guarantees existence even when we cannot find the function.
  2. Local Behavior: The function $g(x)$ is defined implicitly. Compute its first and second derivatives, $g'(0)$ and $g''(0)$, using implicit differentiation.

Exercise 2: The Implicit Function Theorem and Manifolds

The equation $x^2 + y^2 + z^2 - 1 = 0$ defines the **unit sphere**, a fundamental example of a smooth manifold.

  1. Local Solvability: Show that the Implicit Function Theorem allows us to locally solve for $z$ as a function of $(x,y)$ near the point $(0,0,1)$. What geometric insight does this give you about the surface at this point?
  2. Geometric Interpretation of Derivatives: Compute the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at the point $(0,0)$. Explain how the values of these derivatives relate to the tangent plane of the sphere at the point $(0,0,1)$.

Exercise 3: A Hyperboloid of One Sheet

Consider the equation $x^2 + y^2 - z^2 - 1 = 0$, which defines a **hyperboloid of one sheet**.

  1. Applicability of the Theorem: Show that the Implicit Function Theorem can be used to locally solve for $z$ as a function of $(x,y)$ near the point $(1,0,\sqrt{2})$. Why is this point a good choice, and what would happen if we tried to solve for $z$ at a point where $z=0$?
  2. Tangent Space: Using implicit differentiation, compute the partial derivatives $\frac{\partial z}{\partial x}(1,0)$ and $\frac{\partial z}{\partial y}(1,0)$. Explain how these values can be used to describe the tangent plane to the hyperboloid at $(1,0,\sqrt{2})$.
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