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Exploring Behaviors of
Implicit Relations

Learn how to analyze slopes, tangent lines, and critical features of curves defined by implicit equations using implicit differentiation.

AP Calculus AB

What Is an Implicit Relation?

An explicit function isolates one variable: \( y = f(x) \). An implicit relation intertwines both variables in a single equation, like \( x^2 + y^2 = 25 \). We cannot always solve for \( y \) neatly, yet the curve still has well-defined slopes and tangent lines at most points.

Why Do Implicit Relations Matter?

Curves that fail the VLT
Circles, ellipses, and lemniscates are not functions, but calculus still applies locally.
Real-world models
Physics and economics often produce equations where isolating \( y \) is impractical or impossible.
AP Exam staple
Free-response questions frequently ask for tangent lines, critical points, and concavity of implicit curves.

Visualizing \( x^2 + y^2 = 25 \)

x y 5 -5 5 -5 (3, 4) tangent \( \text{slope} = -\tfrac{3}{4} \)

The Core Rule: Implicit Differentiation

Differentiate both sides of the equation with respect to \( x \). Every time you differentiate a term containing \( y \), multiply by \( \frac{dy}{dx} \) via the Chain Rule.

The Chain Rule Connection

Differentiate outer function Multiply by \( \frac{dy}{dx} \) Solve for \( \frac{dy}{dx} \)

When you see \( y^n \), think of it as \( [g(x)]^n \) where \( g(x) = y \). The Chain Rule demands that the derivative of the "inner" function, \( \frac{dy}{dx} \), appears as a multiplier every time.

Explicit vs. Implicit Differentiation

Explicit Implicit
Form \( y = f(x) \) \( F(x, y) = 0 \)
Solve for \( y \)? Yes, always Often impossible
Differentiation Standard rules Chain Rule on every \( y \)
Result for \( \frac{dy}{dx} \) In terms of \( x \) only In terms of both \( x \) and \( y \)
Example \( y = \sqrt{25 - x^2} \) \( x^2 + y^2 = 25 \)

How to Apply Implicit Differentiation

1
Differentiate every term on both sides with respect to \( x \).
2
Apply the Chain Rule to \( y \)-terms: attach \( \frac{dy}{dx} \) each time you differentiate \( y \).
3
Collect all \( \frac{dy}{dx} \) terms on one side of the equation.
4
Factor out \( \frac{dy}{dx} \) and divide to isolate it.
5
Substitute the point \( (x_0, y_0) \) to find the numerical slope, if needed.

Example 1: \( x^2 + y^2 = 25 \)

1
Differentiate: \( 2x + 2y\,\frac{dy}{dx} = 0 \)
2
Isolate: \( 2y\,\frac{dy}{dx} = -2x \)
3
Solve: \( \frac{dy}{dx} = -\frac{x}{y} \)

Example 2: \( x^3 + y^3 = 6xy \)

This is the Folium of Descartes, a classic implicit curve.

1
Differentiate: \( 3x^2 + 3y^2\,\frac{dy}{dx} = 6y + 6x\,\frac{dy}{dx} \)
2
Collect: \( 3y^2\,\frac{dy}{dx} - 6x\,\frac{dy}{dx} = 6y - 3x^2 \)
3
Factor and solve: \( \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x} \)

Example 3: Horizontal & Vertical Tangents

For the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \), find where the tangent is horizontal or vertical.

x y (0, 2) (0, -2) (3, 0) (-3, 0) Horizontal: \( \frac{dy}{dx} = 0 \) Vertical: \( \frac{dy}{dx} \) undefined

Finding \( \frac{d^2y}{dx^2} \) Implicitly

To analyze concavity of implicit curves, differentiate \( \frac{dy}{dx} \) again with respect to \( x \). The result will contain \( \frac{dy}{dx} \), which you substitute back in.

1
Start with \( \frac{dy}{dx} = -\frac{x}{y} \) from \( x^2 + y^2 = 25 \).
2
Differentiate using the Quotient Rule: \( \frac{d^2y}{dx^2} = -\frac{y - x\,\frac{dy}{dx}}{y^2} \)
3
Substitute \( \frac{dy}{dx} = -\frac{x}{y} \): \( \frac{d^2y}{dx^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{25}{y^3} \)

Writing the Tangent Line Equation

Once you have \( \frac{dy}{dx} \) at a specific point \( (x_0, y_0) \), plug into point-slope form:

Example: For \( x^2 + xy + y^2 = 7 \) at the point \( (1, 2) \):

1
Differentiate: \( 2x + y + x\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} = 0 \)
2
Solve: \( \frac{dy}{dx} = -\frac{2x+y}{x+2y} \). At (1,2): \( \frac{dy}{dx} = -\frac{4}{5} \)
3
Tangent line: \( y - 2 = -\frac{4}{5}(x - 1) \)

Behaviors at Special Points

Horizontal Tangent
\( \frac{dy}{dx} = 0 \): the numerator of the derivative equals zero while the denominator does not.
Vertical Tangent
\( \frac{dy}{dx} \) is undefined: the denominator equals zero while the numerator does not.
Singular Point
Both numerator and denominator are zero. The curve may have a cusp, node, or isolated point.

Common Mistakes to Avoid

✕ Forgetting \( \frac{dy}{dx} \)
Writing \( \frac{d}{dx}[y^2] = 2y \) instead of \( 2y\,\frac{dy}{dx} \). Every \( y \)-term needs the chain rule factor.
✕ Dropping the Product Rule
For \( xy \), you must use the product rule: \( \frac{d}{dx}[xy] = y + x\,\frac{dy}{dx} \), not just \( \frac{dy}{dx} \).
✕ Not Verifying the Point
Before substituting into \( \frac{dy}{dx} \), always confirm the point actually satisfies the original equation.

Summary: Key Behaviors

Behavior Condition on \( \frac{dy}{dx} \) What to Check
Horizontal tangent \( \frac{dy}{dx} = 0 \) Numerator = 0, denominator \( \neq 0 \)
Vertical tangent \( \frac{dy}{dx} \) undefined Denominator = 0, numerator \( \neq 0 \)
Concave up \( \frac{d^2y}{dx^2} > 0 \) Differentiate \( \frac{dy}{dx} \) again
Concave down \( \frac{d^2y}{dx^2} < 0 \) Differentiate \( \frac{dy}{dx} \) again
Increasing \( \frac{dy}{dx} > 0 \) Sign of derivative at the point
Decreasing \( \frac{dy}{dx} < 0 \) Sign of derivative at the point

Question 1 of 5

Given \( x^2 + y^2 = 25 \), what is \( \frac{dy}{dx} \) at the point \( (3, -4) \)?

Question 2 of 5

For the equation \( x^2 + xy + y^2 = 7 \), which expression equals \( \frac{dy}{dx} \)?

Question 3 of 5

The curve \( y^2 = 4x \) has a vertical tangent at which point?

Question 4 of 5

For \( x^2 + y^2 = 25 \), the second derivative \( \frac{d^2y}{dx^2} \) equals \( -\frac{25}{y^3} \). At the point \( (3, 4) \), the curve is:

Question 5 of 5

When differentiating the term \( \sin(y) \) with respect to \( x \) in an implicit equation, the result is: