Implicit Functions

When Does an Equation Secretly Define a Function?

Take the familiar equation $x^2 + y^2 = 1$. We know it traces a circle, but let's think about what it means to "solve for $y$." If we try, we get $y = \pm\sqrt{1 - x^2}$ - two branches, not one. That plus-or-minus is the problem.

But here's the key observation: if we zoom in near the point $(0, 1)$ - the top of the circle - we only see one branch. The lower half is far away. Locally, the equation quietly defines $y = \sqrt{1 - x^2}$, a perfectly good function.

Now zoom in near $(1, 0)$ - the rightmost point. Both branches collide there. The curve goes vertical. There's no way to write $y$ as a single-valued function of $x$ near that spot.

So here's the setup: we always start with a known point $(x_0, y_0)$ that already sits on the curve - meaning $F(x_0, y_0) = 0$. The question isn't "does a solution exist somewhere?" It's: can we smoothly track that solution as we nudge $x$ away from $x_0$? Does each nearby $x$ give exactly one nearby $y$?

The answer comes from $F_y$. Write $F(x,y) = x^2 + y^2 - 1$. At the known point $(0, 1)$, we have $F_y = 2y = 2 \neq 0$. That means $F$ is strictly increasing in $y$ there - so if you fix any $x$ near $0$ and slide $y$ up through values near $1$, $F$ crosses zero exactly once. That unique crossing is our implicit function: $y = f(x)$.

At $(1, 0)$, by contrast, $F_y = 2y = 0$. The function $F$ isn't monotone in $y$ there - it bottoms out instead of crossing through. So there's no clean single crossing to track, and the implicit function breaks down.

The visualization below makes this concrete. The background shading shows $|F_y|$: bright green where $|F_y|$ is large (safe zones), red where $|F_y|$ is small (danger zones). Switch between curves to see the pattern repeat.

The Derivative You Can't See

We've established that near a point where $F_y \neq 0$, the equation $F(x,y) = 0$ secretly defines $y$ as a function of $x$. But what's the derivative $dy/dx$ of that hidden function? We can't just differentiate an explicit formula - we might not have one.

The trick is beautiful: since $F(x, y(x)) = 0$ holds for all $x$ near our point, we can differentiate both sides with respect to $x$. The right side gives zero. The left side requires the chain rule - and let's walk through it carefully.

The Derivation

Start with the identity that holds everywhere on the curve:

Differentiate both sides with respect to $x$. On the left, $F$ depends on $x$ in two ways: directly, and through $y$. The chain rule gives:

Since $dx/dx = 1$, this simplifies to:

Now solve for $dy/dx$:

That minus sign isn't arbitrary. Think about it: if increasing $x$ pushes $F$ upward ($F_x > 0$), then $y$ must decrease to compensate and keep $F = 0$. The curve fights back.

Numerical Check

Let's verify this on the unit circle at the point $\left(\frac{1}{2},\, \frac{\sqrt{3}}{2}\right)$. We have $F(x,y) = x^2 + y^2 - 1$, so:

From the explicit formula $y = \sqrt{1-x^2}$, differentiate directly: $dy/dx = -x/\sqrt{1-x^2} = -\tfrac{1/2}{\sqrt{3}/2} = -1/\sqrt{3}$. They match.

The implicit formula gave us the slope without ever solving for $y$. That's the whole point - for complicated curves where you can't solve explicitly, this is the only game in town.

Three Variables - Surfaces

Everything we just did with curves in the plane extends naturally to surfaces in space. An equation $F(x, y, z) = 0$ defines a surface. Near a point where $F_z \neq 0$, we can locally solve for $z = f(x, y)$ - the surface becomes the graph of a function.

The derivation is identical. Differentiate $F(x, y, z(x,y)) = 0$ with respect to $x$ (holding $y$ constant):

And similarly with respect to $y$:

Example: The Unit Sphere

Take $F(x,y,z) = x^2 + y^2 + z^2 - 1$. Then $F_x = 2x$, $F_y = 2y$, $F_z = 2z$, so:

At the point $\left(\tfrac{1}{2},\, \tfrac{1}{2},\, \tfrac{1}{\sqrt{2}}\right)$ (check: $\tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{2} = 1$), we get:

Drag the point on the sphere below. The orange tangent plane shows the local linear approximation $z \approx f(a,b) + z_x(x-a) + z_y(y-b)$. Watch what happens near the equator.

Simultaneous Equations and Jacobians

What if we have two equations relating four variables? For instance:

Two equations, four unknowns. We'd expect to "use up" two degrees of freedom, leaving two free variables. So if we pick $x$ and $y$ as independent, the system should (locally) determine $z$ and $w$ as functions of $x$ and $y$.

But which pairs of variables can serve as "outputs"? This is where Jacobians enter.

The Setup

Take total differentials of both equations:

If we want $z$ and $w$ as functions of $x$ and $y$, we hold $y$ constant ($dy = 0$) and ask: what are $dz$ and $dw$ in terms of $dx$? This gives a $2 \times 2$ linear system:

By Cramer's rule, we can solve for $dz/dx$:

Notice the pattern: the denominator is always the Jacobian of $(F,G)$ with respect to the output variables $(z,w)$. The numerator swaps one output variable for the input variable whose partial you want.

Example

Consider the system at the point $(x, y, z, w) = (1, 1, 1, 4)$:

Compute all needed partials at $(1,1,1,4)$:

The denominator Jacobian:

The numerator for $\partial z/\partial x$:

Therefore:

The Implicit Function Theorem

After four cards of examples, let's step back and see the unified picture. Everything we've done - one equation in two unknowns, one equation in three unknowns, two equations in four unknowns - is a special case of a single, powerful theorem.

The General Statement

Suppose we have $m$ equations in $m + n$ variables:

At a point where these equations hold and the $m \times m$ Jacobian

there exist unique smooth functions $y_1 = f_1(\mathbf{x}), \ldots, y_m = f_m(\mathbf{x})$ defined in a neighborhood of the point, satisfying all $m$ equations.

One condition - a single determinant being nonzero - simultaneously guarantees existence, uniqueness, and smoothness. That's remarkably efficient.

The derivative formula also generalizes. Each partial derivative is a ratio of Jacobians:

Special Cases We've Seen

The pattern is always the same: count equations, count unknowns, identify which variables you want to solve for, check that the Jacobian with respect to those variables is nonzero. If it is, you're guaranteed a smooth local solution, and you can compute all its derivatives by implicit differentiation.