The Implicit Function Theorem

Prerequisites

• Derivatives, Differentials, and the Chain Rule — partial derivatives, chain rule
• Directional Derivatives and the Gradient — gradient perpendicular to level sets

The problem

You have a constraint $x^{3}y+z^2=2$ and you want to compute $\frac{\partial x}{\partial z}$ with $y$ held fixed. To do this, you write $x=x(z)$ and differentiate with the chain rule. But wait — how do you know that the constraint actually defines $x$ as a function of $z$ near your point? Maybe it doesn’t. Maybe there are two values of $x$ for each $z$, or none. You need a guarantee before you can differentiate.

The Implicit Function Theorem is that guarantee. It tells you: under what condition does a constraint equation locally define one variable as a smooth function of the others? And when it does, what is the derivative?

The key idea

A constraint $F(x,y)=c$ defines a curve in the plane. You can write $y=\varphi (x)$ locally — meaning the curve is a graph over the $x$-axis — if and only if the curve is not vertical at your point.

“Not vertical” means: the tangent line to the curve has a nonzero $x$-component. Since the gradient $\nabla F=(F_x,F_y)$ is perpendicular to the curve, the tangent direction is $(-F_y,F_x)$. This has nonzero $x$-component precisely when $F_y\ne 0$.

That’s the whole condition: $F_y\ne 0$ means the curve isn’t vertical, which means $y$ is locally a function of $x$.

Why “not vertical” is the right condition

Consider $x^2+y^2=1$ (the unit circle).

Near the top $(0,1)$: the tangent is horizontal. The curve passes the vertical line test — for each $x$ near 0, there is exactly one $y$ near 1 on the curve. Indeed $y=\sqrt{1-x^2}$ works. Here $F_y=2y=2\ne 0$. The IFT applies.

Near the rightmost point $(1,0)$: the tangent is vertical. For $x$ slightly less than 1, the circle has two $y$-values ($\pm \sqrt{1-x^2}$). For $x$ slightly greater than 1, there are none. No function $y=\varphi (x)$ can describe the curve here. And indeed $F_y=2y=0$. The IFT correctly refuses to apply.

But $F_x=2x=2\ne 0$ at $(1,0)$, so you can solve for $x$ as a function of $y$ there: $x=\sqrt{1-y^2}$. The curve is horizontal at that point — it’s a graph over the $y$-axis, just not over the $x$-axis. Which variable you can solve for depends on which partial is nonzero.

The theorem (2D)

In words: If you have an equation $F(x,y)=c$ and the function $F$ is smooth, then near any point where $F$ changes in the $y$-direction (i.e., $F_y\ne 0$), the equation locally defines $y$ as a smooth function of $x$. Moreover, you can compute its derivative by implicit differentiation.

Hypothesis: Let $F:U\subset {\mathbb{R}}^2\to \mathbb{R}$ be $C^1$, and let $(x_0,y_0)$ satisfy:

1. $F(x_0,y_0)=c$
2. $F_y(x_0,y_0)\ne 0$

Thesis: There exist open intervals $I\ni x_0$ and $J\ni y_0$, and a unique $C^1$ function $\varphi :I\to J$ such that: $F(x,y)=c⟺y=\varphi (x)\text{for all }(x,y)\in I\times J$ with derivative: ${\varphi }^{\prime }(x)=-\frac{F_x(x,\varphi (x))}{F_y(x,\varphi (x))}$

The $C^1$ condition on $F$ matters: the proof needs $F_y\ne 0$ on a neighborhood (not just at the point) to get a uniform contraction estimate, and continuity of $F_y$ is what guarantees nonvanishing persists nearby.

The theorem (3D — one constraint)

In words: If you have a constraint $G(x,y,z)=c$ and $G$ responds to changes in $z$ at your point ($G_z\ne 0$), then the constraint locally defines $z$ as a smooth function of $x$ and $y$. The surface is locally a graph over the $(x,y)$-plane, and its slopes are computable by implicit differentiation.

Hypothesis: Let $G:U\subset {\mathbb{R}}^3\to \mathbb{R}$ be $C^1$, with $G(x_0,y_0,z_0)=c$ and $G_z(x_0,y_0,z_0)\ne 0$.

Thesis: There exists a neighborhood $V$ of $(x_0,y_0)$ and a unique $C^1$ function $\psi :V\to \mathbb{R}$ such that $G(x,y,\psi (x,y))=c$, with: ${\psi }_x=-\frac{G_x}{G_z},{\psi }_y=-\frac{G_y}{G_z}$

Geometrically: $G_z\ne 0$ means the tangent plane to the surface is not vertical (not parallel to the $z$-axis), so the surface is locally a graph over the $(x,y)$-plane.

Worked example: $x^{3}y+z^2=2$

Goal: Find $\frac{\partial x}{\partial z}$ with $y$ held fixed at the point $(1,1,1)$.

Step 1 — Verify the constraint is satisfied: $1^3\cdot 1+1^2=2✓$

Step 2 — Check the IFT hypothesis. We want to solve for $x$, so we need $G_x\ne 0$: $G_x=3x^{2}y=3(1{)}^2(1)=3\ne 0✓$

The IFT guarantees that $x=x(y,z)$ exists as a smooth function near this point.

Step 3 — Differentiate the constraint with $y$ held fixed (so $dy=0$): $G_x\frac{\partial x}{\partial z}+G_z=0$ $3x^{2}y\cdot \frac{\partial x}{\partial z}+2z=0$

Step 4 — Solve: $\frac{\partial x}{\partial z}{|}_{(1,1,1)}=-\frac{G_z}{G_x}=-\frac{2z}{3x^{2}y}=-\frac{2}{3}$

Why this justifies “implicit differentiation”

The derivative formula ${\varphi }^{\prime }=-F_x/F_y$ is exactly what students compute when they “differentiate both sides” of $F(x,y)=c$. The chain rule gives $F_x+F_y{\varphi }^{\prime }=0$, and you solve.

But the chain rule requires that $\varphi$ exists and is differentiable — both of which are assumptions, not free. The IFT is the theorem that provides both. Without it, you’re computing the derivative of a function that might not exist.

A subtlety: sufficiency, not necessity

The condition $F_y\ne 0$ is sufficient for $y=\varphi (x)$ to exist locally. It is not necessary.

Example: $F(x,y)=y^3$, level set $F=0$. At the origin, $F_y=3y^2=0$ — the IFT does not apply. But the level set is just $\{y=0\}$, a perfectly smooth graph. The solution exists; the IFT simply can’t detect it.

The IFT works generically. The cases where it fails but a solution still exists (cusps, flat tangency) require ad hoc analysis.

Going deeper

Equivalence with the Inverse Function Theorem

The IFT and the Inverse Function Theorem are the same theorem in two presentations.

Inverse FT $\Rightarrow$ Implicit FT: Given $F(x,y)=c$ with $F_y\ne 0$, define $\Phi (x,y)=(x,F(x,y))$. Its Jacobian determinant is $F_y\ne 0$, so by the Inverse FT, $\Phi$ is locally invertible. The level set $F=c$ corresponds to $v=c$ in the new coordinates, giving $\varphi (x):={\Phi }^{-1}(x,c{)}_2$.

Implicit FT $\Rightarrow$ Inverse FT: Given $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ with $Df(p)$ invertible, define $F(x,y)=f(x)-y$. Then $F_x=Df$ is invertible, so the IFT solves $F=0$ as $x=g(y)$, which is the local inverse.

Where differentiability comes from

Continuous existence of $\varphi$ follows from monotonicity of $F$ in $y$ (a topological fact — intermediate value theorem). Differentiability is harder: it comes from the fact that $F$‘s linear approximation is invertible. The argument linearizes the constraint, solves for the linear part, and shows the remainder is $o(|h|)$ via a Lipschitz bound from the contraction mapping.

The $C^1$ hypothesis ensures $F_y\ne 0$ on a neighborhood (not just at the point), which makes the contraction uniform.

Questions to sit with

1. On the sphere $x^2+y^2+z^2=1$, at the “equator” point $(1,0,0)$: $G_x=2\ne 0$, $G_y=0$, $G_z=0$. Which variable(s) can you solve for? What surface patch do you get?

2. The formula ${\varphi }^{\prime }=-F_x/F_y$ blows up when $F_y\to 0$ — the tangent becoming vertical. Is this analogous to the single-variable phenomenon where a function’s inverse fails to be differentiable at a point where $f^{\prime }=0$?

See also

• Derivatives, Differentials, and the Chain Rule — the chain rule that the IFT justifies applying
• Directional Derivatives and the Gradient — gradient perpendicular to level sets
• Lagrange Multipliers — uses the IFT implicitly (constraint defines a surface)