Implicit Relations

An implicit relation defines a connection between variables (typically $x$ and $y$ ) through an equation $F (x, y) = 0$ that is not solved for one variable in terms of the other. Key result: even without an explicit formula $y = f (x)$ , the slope $d y / d x$ can be found by implicit differentiation. Applies wherever curves resist algebraic solving — circles, ellipses, folium curves, thermodynamic state equations. Formalized through 17th–18th century calculus (Descartes, Leibniz, Newton) and rigorized by the Implicit Function Theorem (Dini, 1870s).

An implicit relation is an equation of the form $F (x, y) = 0$ that constrains the pairs $(x, y)$ lying on a curve, without expressing $y$ as an explicit function of $x$ . This matters because most curves of interest — circles, ellipses, lemniscates, and the level sets of any two-variable function — cannot be cleanly solved for $y$ , yet we still need their slopes, tangents, and shapes. The central technique is implicit differentiation: differentiate both sides with respect to $x$ , treating $y$ as a hidden function of $x$ and applying the chain rule. The surprising fact is that you can compute the exact slope at a point on a curve you can never algebraically untangle — for instance the tangent to $x^{3} + y^{3} = 3 x y$ (the folium of Descartes) — using only the equation itself. Implicit relations are the natural language for level curves and underpin much of multivariable calculus.

The Core Insight

A single equation in two variables generally defines a curve, not a function. Locally — near most points — that curve behaves like a function $y = f (x)$ even when no formula exists. Differentiating the constraint $F (x, y) = 0$ as an identity lets us extract the slope of that hidden function directly, because the chain rule converts " $y$ depends on $x$ " into algebra we can solve for $d y / d x$ .

Formal Statement

Explicit vs. Implicit Relation

An explicit function expresses one variable directly:

y = f (x)

⟳this is the definition — the output

y

is isolated on the left, computed by feeding

x

into the rule

f

An implicit relation is an equation

F (x, y) = 0

⟳any equation in

x

and

y

can be moved to one side; the zero set

{(x, y) : F (x, y) = 0}

is the curve it defines.

Here $F$ is a function of two variables, and the relation is the set of points satisfying it. Symbols: $x, y$ are the coordinate variables; $F$ is the constraint function; $F = 0$ is its zero level set.

Recognizing the forms

The unit circle is implicit: $x^{2} + y^{2} - 1 = 0$ . Solving for $y$ gives two explicit functions, $y = 1 - x^{2}$ (upper half) and $y = - 1 - x^{2}$ (lower half). One implicit relation often packs several explicit branches.

Implicit Differentiation

If $F (x, y) = 0$ defines $y$ as a differentiable function of $x$ near a point, then differentiating with respect to $x$ and applying the chain rule gives:

\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{d y}{d x} = 0

⟳differentiate

F (x, y (x)) = 0

; the term in

x

gives

\partial F / \partial x

, the term in

y

gives

\partial F / \partial y

times

d y / d x

by the chain rule.

Solving for the slope:

\frac{d y}{d x} = - \frac{\partial F / \partial x}{\partial F / \partial y}, \frac{\partial F}{\partial y} \neq = 0

⟳isolate

d y / d x

in equation [3] by subtracting and dividing by

\partial F / \partial y

Implicit Function Theorem (introductory form)

If $F$ is continuously differentiable near a point $(a, b)$ with $F (a, b) = 0$ and $\frac{\partial F}{\partial y} (a, b) \neq = 0$ , then near $(a, b)$ the relation $F (x, y) = 0$ determines a unique differentiable function $y = f (x)$ with $f (a) = b$ , and its derivative is given by equation [4].

Why It Holds / Derivation

Implicit Differentiation via the Chain Rule

Setup. Assume that near the point of interest the curve $F (x, y) = 0$ coincides with the graph of some function $y = y (x)$ , so that the identity $F (x, y (x)) = 0$ holds for all nearby $x$ .

Differentiate both sides w.r.t. $x$ . The right side is constant, so its derivative is $0$ . The left side is a composite of two variables both depending (the first trivially) on $x$ .

Apply the multivariable chain rule. For $G (x) = F (x, y (x))$ ,

\frac{d G}{d x} = \frac{\partial F}{\partial x} \frac{d x}{d x} + \frac{\partial F}{\partial y} \frac{d y}{d x}

⟳total derivative = sum of partial derivatives times the derivative of each argument;

d x / d x = 1

Set equal to zero. Since $G (x) \equiv 0$ , we get $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{d y}{d x} = 0$ , which is equation [3].

Solve. Provided $\partial F / \partial y \neq = 0$ , rearrange to obtain equation [4]. The condition $\partial F / \partial y \neq = 0$ is exactly what guarantees the curve is not vertical at that point, so a function $y$ of $x$ exists locally. $■$

Ulisse Dini | 1845–1918

The Italian analyst Ulisse Dini gave the modern rigorous statement and proof of the Implicit Function Theorem in his lectures of the 1870s, establishing exactly when an equation $F (x, y) = 0$ can be solved for $y$ near a point. The result — and the careful hypothesis $\partial F / \partial y \neq = 0$ — turned the centuries-old practical trick of implicit differentiation into a guaranteed theorem of analysis. His name still attaches to the theorem in Italian and French mathematical traditions.

Building Intuition

Picture the equation $F (x, y) = 0$ as the shoreline where a landscape's height $F$ exactly hits sea level. Walking along the shore, you never go up or down — your net change in height is zero. Equation [3] is just this statement: as $x$ changes, the height change from moving east ( $\partial F / \partial x$ ) must be cancelled by the height change from moving north ( $\partial F / \partial y \cdot d y / d x$ ). The slope $d y / d x$ is whatever direction keeps you exactly on the shoreline. When the shoreline runs straight north–south (a vertical tangent), $\partial F / \partial y = 0$ and you can't describe $y$ as a function of $x$ there — equation [4] correctly blows up.

The unit circle as an implicit relation, with tangent slope from implicit differentiation

Worked Examples

Slope on the unit circle

Given:

x^{2} + y^{2} = 1

. Find

d y / d x

and evaluate at

P = (\frac{2}{2}, \frac{2}{2})

. Steps:

Write $F = x^{2} + y^{2} - 1$ . Differentiate term by term w.r.t. $x$ :

2 x + 2 y \frac{d y}{d x} = 0

⟳

\frac{d}{d x} x^{2} = 2 x

;

\frac{d}{d x} y^{2} = 2 y \frac{d y}{d x}

by the chain rule;

\frac{d}{d x} 1 = 0

Solve: $\frac{d y}{d x} = - \frac{x}{y}$ (matches equation [4] with $\partial F / \partial x = 2 x$ , $\partial F / \partial y = 2 y$ ).

Substitute $x = y = \frac{2}{2}$ : $\frac{d y}{d x} = - \frac{2 /2}{2 /2} = - 1$ .

\frac{d y}{d x}_{P} = - 1 (dimensionless slope)

This is perpendicular to the radius, as geometry predicts.

Folium of Descartes — a genuinely implicit curve

Given:

x^{3} + y^{3} = 3 x y

. Find the tangent slope at

(\frac{3}{2}, \frac{3}{2})

. Steps:

Differentiate both sides w.r.t. $x$ (product rule on $3 x y$ ):

3 x^{2} + 3 y^{2} \frac{d y}{d x} = 3 y + 3 x \frac{d y}{d x}

⟳

\frac{d}{d x} y^{3} = 3 y^{2} y^{'}

;

\frac{d}{d x} (3 x y) = 3 y + 3 x y^{'}

by the product rule.

Collect $d y / d x$ terms:

\frac{d y}{d x} (3 y^{2} - 3 x) = 3 y - 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{y - x ^{2}}{y ^{2} - x}

⟳move all

y^{'}

terms left, factor, divide; cancel the common factor

3

Substitute $x = y = \frac{3}{2}$ : numerator $= \frac{3}{2} - \frac{9}{4} = - \frac{3}{4}$ ; denominator $= \frac{9}{4} - \frac{3}{2} = \frac{3}{4}$ .

\frac{d y}{d x} = \frac{- 3/4}{3/4} = - 1 (dimensionless)

There is no explicit $y = f (x)$ formula here that is practical, yet the slope is exact.

Common Misconceptions

"Just solve for $y$ first." For most implicit relations you cannot —

x^{3} + y^{3} = 3 x y

has no nice closed form, and even

x^{2} + y^{2} = 1

splits into two functions. Implicit differentiation exists precisely so you don't have to solve. Also: when you differentiate a term containing

y

, you must attach

\frac{d y}{d x}

(the chain rule). Forgetting it is the single most common error. $d y / d x$ may depend on both $x$ and $y$ . Unlike explicit derivatives, the answer in equation [4] usually involves

y

too. To get a number you need a point

(x, y)

on the curve, not just an

x

-value. Plugging in an

x

alone is meaningless because the same

x

can correspond to several

y

's.

When the theorem fails

Consider $x^{2} + y^{2} = 0$ . The only real solution is the single point $(0, 0)$ — not a curve at all, so no function $y = f (x)$ exists in any neighborhood. Here $\partial F / \partial y = 2 y = 0$ at the only point, violating the Implicit Function Theorem's hypothesis. Likewise on $x^{2} + y^{2} = 1$ at $(\pm 1, 0)$ , $\partial F / \partial y = 0$ and the tangent is vertical: $d y / d x$ is undefined and no single function $y (x)$ describes the curve there.

Connections

Implicit relations sit at a crossroads of several ideas. The technique relies entirely on the Chain Rule and the Product Rule. The condition $\partial F / \partial y \neq = 0$ introduces Partial Derivatives and previews the full Implicit Function Theorem of multivariable calculus. Curves $F (x, y) = 0$ are exactly the level curves studied in Level Sets and Contour Lines, and the gradient $(\partial F / \partial x, \partial F / \partial y)$ is always perpendicular to them, linking to Gradient Vector. When a curve is instead given by $x (t), y (t)$ , the analogous slope comes from Parametric Differentiation.

Real-World Applications

Related-rates problems (physics & engineering): a sliding ladder $x^{2} + y^{2} = L^{2}$ relates the speeds of its ends via implicit differentiation in time.
Thermodynamics: the van der Waals equation of state $(P + \frac{a}{V ^{2}}) (V - b) = R T$ is implicit in $P, V, T$ ; partial derivatives extract compressibility.
Economics — indifference curves: a utility level $U (x, y) = c$ defines an implicit curve whose slope (the marginal rate of substitution) follows from equation [4].
Computer graphics: implicit surfaces $F (x, y, z) = 0$ (metaballs, blobby models) are rendered via ray–level-set intersections.
GPS and surveying: intersecting distance constraints (circles, hyperbolas) are implicit relations solved jointly.
Optics: wavefronts are level sets $ϕ (x, y) = const$ ; rays travel along the gradient, perpendicular to these implicit curves.
Mechanical linkages & robotics: joint constraints are implicit equations whose differentiation gives velocity relationships (the Jacobian).

Historical Development

1637 — René Descartes introduces coordinate geometry in La Géométrie; the folium $x^{3} + y^{3} = 3 x y$ becomes a famous test curve.
1660s–1680s — Newton and Leibniz independently develop calculus; both differentiate equations not solved for one variable.
1684 — Leibniz publishes differentiation rules (Nova Methodus), making chain-rule-style manipulation systematic.
1696 — L'Hôpital's textbook Analyse des Infiniment Petits presents tangent-finding on implicit curves to a broad audience.
1740s — Euler formalizes functions of several variables, clarifying partial derivatives that underlie $\partial F / \partial y$ .
1870s — Ulisse Dini states and proves the rigorous Implicit Function Theorem, fixing exact existence conditions.
Late 19th c. — The theorem becomes a cornerstone of real analysis and, generalized, of differential geometry.
20th c. — Extensions to Banach spaces and the inverse function theorem cement its role across mathematics and physics.

Subtleties & Limits

The local Implicit Function Theorem guarantees a function $y = f (x)$ only in a small neighborhood where $\partial F / \partial y \neq = 0$ . Globally, the same curve may fail to be a function (it loops, self-intersects, or turns vertical), and there is no simple universal criterion telling you how large the valid neighborhood is. Determining the global structure of an implicit curve — its number of branches, singular points (where both partials vanish), and connectivity — is the subject of algebraic geometry and remains computationally hard for high-degree polynomial relations. Even deciding whether $F (x, y) = 0$ has real solutions at all can be subtle.