The Implicit Function Theorem

The equation x^2 + y^2 = 1 draws a circle, and a circle is emphatically not the graph of a function: the vertical line through x = 0.5 hits it twice, at a top point and a bottom point. And yet — cover the bottom half with your hand and the top arc is a perfectly good function y = \sqrt{1 - x^2}. Zoom in on almost any point of the circle and it looks like the graph of a tidy little function y = g(x), even though the whole curve is not.

This is the everyday miracle the Implicit Function Theorem makes precise. An equation F(x, y) = 0 ties x and y together implicitly; the theorem tells you exactly when you may, near a given point, untie it into an explicit y = g(x) — and it does so even when no formula for g exists. It then hands you the derivative g'(x) for free, by differentiating the constraint you already have.

Constraint curves and surfaces are everywhere: an equation of state PV = nRT relating pressure, volume and temperature; an indifference curve in economics; a level set of energy in mechanics; the solution set of any system of equations. The theorem is what lets you treat one variable as a smooth function of the others along such a set — and it is, remarkably, the same theorem as the Inverse Function Theorem seen from a different chair.

The criterion: one partial derivative must not vanish

When can F(x, y) = 0 be solved for y near a point (a, b) that satisfies it? The obstruction is a vertical tangent — the moment the curve turns straight up, a single x owns a whole range of y's and no function can cope. A vertical tangent is precisely where the rate of change of F in the y-direction dies. So the criterion is:

\frac{\partial F}{\partial y}(a, b) \ne 0.

That single nonzero partial derivative is the whole hypothesis. In the general m-equation version you solve for a block of variables and the condition becomes "the corresponding partial Jacobian block is invertible" — the direct heir of the inverse-function determinant test.

Let F : \mathbb{R}^{n} \times \mathbb{R}^{m} \to \mathbb{R}^{m} be C^1, with F(a, b) = 0, and suppose the partial Jacobian in the y-variables, D_y F(a, b), is invertible. Then:

near a there is a unique C^1 map g with g(a) = b and F\big(x, g(x)\big) = 0 for all nearby x — the constraint is locally the graph of y = g(x);
its derivative is got by implicit differentiation, no formula for g required: g'(x) = -\big[D_y F\big]^{-1} D_x F. In the scalar case, \dfrac{dy}{dx} = -\dfrac{\partial F/\partial x}{\partial F/\partial y}.

Ride around the circle and watch the criterion switch

The unit circle is F(x, y) = x^2 + y^2 - 1 = 0, so \partial F/\partial y = 2y. Move the point around and read the tangent line. Almost everywhere it is tilted, the neighbourhood is the graph of some y = g(x), and the theorem's slope formula gives \dfrac{dy}{dx} = -\dfrac{2x}{2y} = -\dfrac{x}{y}.

Steer the point to the far right (1, 0) or far left (-1, 0). There y = 0, so \partial F/\partial y = 0, the slope formula divides by zero, and the tangent stands bolt upright — no local y = g(x) exists. But notice all is not lost: at those very points \partial F/\partial x = 2x \ne 0, so you can instead solve for x as a function of y. The theorem does not say "unsolvable" — it says "solve for the variable whose partial survives."

Worked implicit differentiation

The folium of Descartes, F(x, y) = x^3 + y^3 - 6xy = 0. There is no clean y = g(x), but you never need one. The partials are

\frac{\partial F}{\partial x} = 3x^2 - 6y, \qquad \frac{\partial F}{\partial y} = 3y^2 - 6x,

so wherever 3y^2 - 6x \ne 0 the theorem gives the honest slope of the curve,

\frac{dy}{dx} = -\frac{3x^2 - 6y}{3y^2 - 6x} = -\frac{x^2 - 2y}{y^2 - 2x}.

At the point (3, 3) (which lies on the curve, since 27 + 27 - 54 = 0) this is -\dfrac{9 - 6}{9 - 6} = -1 — the tangent there has slope -1, found without ever isolating y. That is the practical punch of the theorem: a derivative you can compute for a function you cannot write down.

The same trick powers transcendental ties like y\,e^{y} = x, which cannot be solved for y in elementary terms. Here F = y e^y - x, F_x = -1, F_y = e^y(1 + y), and \dfrac{dy}{dx} = \dfrac{1}{e^y(1 + y)} — a perfectly usable slope for a function with no closed form (it is the Lambert-W function in disguise).

Two errors to guard against.

A vanishing partial is not a dead end. When \partial F/\partial y = 0 you may not solve for y there — but the constraint curve still exists, and if some other partial is nonzero you solve for that variable instead. At (1, 0) on the circle you cannot get y(x), yet x = \sqrt{1 - y^2} is a fine local function of y. The theorem only fails at a genuinely singular point, where every first partial vanishes — like the self-crossing origin of the folium, where the curve has two branches and no neighbourhood is any single graph.
The graph is local, not global. On the circle the theorem near (0, 1) gives the upper semicircle only; it knows nothing of the lower arc. As with the inverse function theorem, "there is a function" means "on some neighbourhood." Indeed the two theorems are logically equivalent — apply the inverse function theorem to the map (x, y) \mapsto (x, F(x, y)) and the implicit function g falls straight out. They are one insight told two ways.