Mathwords: Implicit Function or Relation

Key Formula

F(x, y) = 0

Where:

$F$ = A function of two variables that defines the relationship
$x$ = Typically the independent variable
$y$ = Typically the dependent variable, not solved for explicitly

Worked Example

Problem: The equation

x^2 + y^2 = 25

defines a circle implicitly. Find the slope of the tangent line at the point

(3, 4)

using implicit differentiation.

Step 1: Recognize that the equation is implicit because

y

is not isolated. Rewrite it in the standard implicit form.

x^2 + y^2 - 25 = 0

Step 2: Differentiate both sides of the original equation with respect to

x

, treating

y

as a function of

x

.

\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)

Step 3: Apply the power rule and chain rule. The derivative of

y^2

requires the chain rule because

y

depends on

x

.

2x + 2y\,\frac{dy}{dx} = 0

Step 4: Solve for

\frac{dy}{dx}

by isolating it on one side.

\frac{dy}{dx} = -\frac{x}{y}

Step 5: Substitute the point

(3, 4)

to find the slope at that location.

\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}

Answer: The slope of the tangent line at

(3, 4)

is

-\dfrac{3}{4}

.

Another Example

This example uses the relation from the formal definition and involves the product rule for the $xy$ term, showing how implicit differentiation handles more complex expressions compared to the simpler circle in the first example.

Problem: Given the implicit relation

x^2 + xy - y^2 = 1

, verify that the point

(1, 1)

lies on the curve, and then find

\frac{dy}{dx}

at that point.

Step 1: Check that

(1, 1)

satisfies the equation by substituting

x = 1

and

y = 1

.

(1)^2 + (1)(1) - (1)^2 = 1 + 1 - 1 = 1 \checkmark

Step 2: Differentiate every term with respect to

x

. The term

xy

requires the product rule.

2x + \left(y + x\,\frac{dy}{dx}\right) - 2y\,\frac{dy}{dx} = 0

Step 3: Collect all terms containing

\frac{dy}{dx}

on one side.

\frac{dy}{dx}(x - 2y) = -2x - y

Step 4: Solve for

\frac{dy}{dx}

and substitute the point

(1, 1)

.

\frac{dy}{dx} = \frac{-2x - y}{x - 2y} = \frac{-2(1) - 1}{1 - 2(1)} = \frac{-3}{-1} = 3

Answer: The derivative

\frac{dy}{dx}

at

(1, 1)

is

3

.

Frequently Asked Questions

What is the difference between an implicit and explicit function?

An explicit function directly gives the dependent variable in terms of the independent variable, like

y = 3x + 5

. An implicit function or relation leaves the variables mixed together, like

x^2 + y^2 = 25

. Sometimes you can rearrange an implicit equation into explicit form, but not always — a circle, for instance, cannot be written as a single explicit function of

x

because each

x

-value may correspond to two

y

-values.

How do you find the derivative of an implicit function?

You use implicit differentiation: differentiate both sides of the equation with respect to

x

, applying the chain rule whenever you encounter

y

(since

y

depends on

x

). This introduces

\frac{dy}{dx}

terms, which you then solve for algebraically. The result typically involves both

x

and

y

.

Is every implicit equation a function?

No. An implicit equation defines a relation, but it is only a function if each input

x

produces exactly one output

y

. For example,

x^2 + y^2 = 25

is a relation (a circle) that fails the vertical line test, so it is not a function. However, near most points on the curve, you can locally treat

y

as a function of

x

.

Implicit Function/Relation vs. Explicit Function

	Implicit Function/Relation	Explicit Function
Form	$F(x, y) = 0$ , variables mixed together	$y = f(x)$ , dependent variable isolated
Example	$x^2 + y^2 = 25$	$y = \sqrt{25 - x^2}$
Differentiation	Requires implicit differentiation (chain rule on $y$ )	Standard differentiation rules apply directly
Always a function?	No — may be a relation with multiple $y$ -values per $x$	Yes — each $x$ maps to exactly one $y$
When to use	When isolating $y$ is difficult or impossible	When $y$ can be cleanly written in terms of $x$

Why It Matters

Many curves studied in precalculus and calculus — circles, ellipses, hyperbolas, and more — are naturally described by implicit equations that cannot be neatly solved for

y

. In calculus, implicit differentiation is a core technique for finding slopes and tangent lines on these curves. Beyond math class, implicit relations appear in physics (equations of state), economics (indifference curves), and engineering (constraint equations).

Common Mistakes

Mistake: Forgetting to apply the chain rule to

y

terms during implicit differentiation.

Correction: Every time you differentiate a term containing

y

with respect to

x

, you must multiply by

\frac{dy}{dx}

because

y

is itself a function of

x

. For example,

\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}

, not just

2y

.

Mistake: Assuming an implicit equation always defines a function.

Correction: An implicit equation defines a relation, which may fail the vertical line test. For instance,

x^2 + y^2 = 25

gives two

y

-values for most

x

-values, so it is a relation, not a function. You can often restrict the domain to get a function (e.g., taking only the upper semicircle).

Related Terms

Explicit Function — Dependent variable isolated; contrast to implicit
Implicit Differentiation — Technique for finding derivatives of implicit relations
Function — Maps each input to exactly one output
Relation — General pairing of inputs and outputs
Dependent Variable — The variable whose value depends on another
Independent Variable — The input variable in a function or relation
Equation — Statement of equality that defines the relation