AP Syllabus focus:
‘Techniques for analyzing derivatives, such as using sign charts and classifying extrema, can be extended to curves defined implicitly rather than explicitly as y=f(x).’
Analyzing implicitly defined curves requires extending familiar derivative-based reasoning to relations where y is not isolated. Students apply derivative tests to study extrema, behavior, and overall structure.
Applying Derivative Tests to Implicit Curves
Understanding Implicit Curves and Derivatives
Implicit curves appear when an equation involves x and y together without solving for y in terms of x.

This graphic shows a family of curves defined implicitly by an equation of the form , displayed with a slope field to illustrate directions of . The image highlights that implicit curves can behave locally like functions of with definable tangent directions. The figure includes more detail than required but reinforces how implicit derivatives guide local curve behavior. Source.
Because such relations may represent multiple branches or fail the vertical line test, analyzing their behavior requires differentiating implicitly to obtain derivative expressions that still describe local behavior.
Implicit Curve: A graph defined by an equation involving x and y where y is not expressed explicitly as a function of x.
Before applying derivative tests, the first step is always finding using implicit differentiation. This provides access to the slope of the tangent line, which is essential for determining extrema or behavior along the curve.

This diagram shows a curve and its tangent line at a specific point, illustrating how the derivative gives the instantaneous slope at that location. The tangent line approximates local behavior of the curve. Although shown for an explicit function, the same concept applies to implicitly defined curves. Source.
Once the first derivative is known, students can use standard tools—critical points, sign charts, and concavity tests—adapted to work with expressions involving both variables.
First Derivative and Critical Points on Implicit Curves
A critical point of an implicit curve occurs exactly where the derivative equals zero or does not exist, as long as the point lies on the curve itself. These points often represent potential local maxima, minima, or directional turning points on the implicit relation.
Critical Point: A point on a curve where or does not exist, provided the point satisfies the implicit equation.
To locate critical points on an implicit curve:
Solve the implicit equation to find coordinate pairs that lie on the relation.
Compute the first derivative using implicit differentiation.
Identify where the numerator of is zero (horizontal tangents).
Identify where the denominator of is zero (vertical tangents), as long as the point is actually on the curve.
Recognize that both types of points may contribute to the curve’s overall geometric behavior.
A horizontal tangent, where , typically signals a potential local extremum in the y-direction. A vertical tangent shows undefined slope but may also correspond to crucial geometric transitions when analyzing the relationship.
Using Sign Charts for Implicit Derivatives
Although an implicit curve is not necessarily the graph of a single-valued function y=f(x), derivative tests still depend on the behavior of as x changes. Sign charts organize how the sign of the derivative varies along portions of the curve.
To analyze derivative signs for implicit curves:
Express in simplified form involving x and y.
Identify critical points and any discontinuities.
Evaluate the sign of on portions of the curve by substituting representative points or analyzing component expressions.
Use the resulting sign information to determine whether the curve rises or falls locally as you move along increasing x.
Because implicit curves may have multiple y-values for the same x, sign charts might require analysis across separate branches, emphasizing attention to domain segments where the relation behaves like a function.
Applying Extrema Classification to Implicit Relations
After locating critical points, students classify them using tools identical to those applied to explicit functions. The goal is to determine whether each critical point is a local maximum, local minimum, or neither, based on changes in derivative sign or concavity.
Two primary tests apply:
First Derivative Test: Examine the sign of immediately before and after a critical point along each branch.
Second Derivative Test: Compute and analyze using implicit differentiation to determine concavity if applicable.
= Independent variable
= Dependent variable expressed implicitly
Once the second derivative is computed, classify extrema using concavity criteria:
Concave up near the point and suggests a local minimum.
Concave down near the point and suggests a local maximum.
Indeterminate concavity requires returning to the First Derivative Test.

This figure illustrates how combinations of monotonicity and concavity determine the presence or absence of local extrema. The sign of indicates whether a function is rising or falling, while the sign of shows whether the curve bends upward or downward. The diagram includes more detail than required for this subsubtopic but reinforces how derivative signs classify extrema for both explicit and implicit curves. Source.
A normal descriptive sentence is required here before the next section, ensuring smooth conceptual flow for students transitioning between tests.
Interpreting Derivative Tests on Implicit Curves
Derivative tests for implicit curves help reveal geometric structure that may be hidden when the relation resists solving for y explicitly. Students use these tests to:
Identify where the curve turns, bends, or changes behavior.
Understand how multiple branches of an implicit relation connect or diverge.
Justify claims about maxima, minima, and transitions in slope using derivative-based reasoning.
Extend familiar calculus tools—even sign charts, extrema classifications, and concavity tests—to a broader class of curves than explicit functions alone.
These derivative-based insights ensure that implicit curves can be analyzed with the same rigor and structure expected throughout AP Calculus AB.
FAQ
A local extremum in y depends only on how y changes with respect to x along a branch of the curve. Even if the entire relation is not a function, each branch may behave locally like one.
A point may still be a turning point in the y-direction, provided dy/dx changes sign along that branch.
Because an implicit relation can produce several y-values for a single x-value, each branch may have a different slope. This means the relation can have multiple tangents above the same x-coordinate.
To analyse them, treat each branch separately when computing dy/dx and when applying derivative tests.
Sign charts rely on examining dy/dx as x increases, but implicit curves can loop or fold over themselves, so separate branches must be considered.
Useful strategies include:
• Splitting the curve into distinct branches.
• Testing points that lie on the curve, not just arbitrary x-values.
Second derivatives for implicit relations include x, y and dy/dx, so the expression can be more complex and may differ between branches.
This means concavity may vary sharply across the curve, and it is possible for d2y/dx2 to change sign even when the curve appears smooth, depending on how y varies with x implicitly.
A vertical tangent indicates that dy/dx is undefined, not necessarily that an extremum occurs. However, in rare cases, the curve may switch direction in x rather than y.
To determine whether a local extremum in y exists, examine the behaviour of y along the curve near the point. Only a change in the sign of dy/dx across the point suggests an extremum in the y-direction.
Practice Questions
The curve is defined implicitly by the relation
x² + xy + y² = 7.
(a) Find the x-coordinate(s) of any point(s) on the curve where the tangent is horizontal.
(1–3 marks)
Question 1 (1–3 marks)
• 1 mark: Correctly differentiate the relation implicitly to obtain an expression for dy/dx.
• 1 mark: Correctly set the numerator of dy/dx equal to zero to locate horizontal tangents.
• 1 mark: Correctly identify the x-coordinate(s) of points on the curve that satisfy both the derivative condition and the original relation.
Consider the implicit curve defined by
x³ + y³ = 6xy.
(a) Show that dy/dx can be written as (2y – x²)/(3x² – 2x).
(b) Determine all points on the curve where the tangent line is vertical.
(c) Using derivative tests, decide whether any of these points could be a local extremum of y in terms of x. Justify your reasoning.
(4–6 marks)
Question 2 (4–6 marks)
Part (a)
• 1 mark: Differentiates both sides correctly using implicit differentiation.
• 1 mark: Correctly rearranges to obtain the required expression for dy/dx.
Part (b)
• 1 mark: Correctly identifies that vertical tangents occur when the denominator of dy/dx equals zero, provided the point lies on the curve.
• 1 mark: Correctly determines all such points that also satisfy the original implicit equation.
Part (c)
• 1 mark: Attempts to apply a derivative test (first or second derivative) to classify behaviour.
• 1 mark: Provides a correct justification explaining why the identified points either could or could not be local extrema of y in terms of x, using sign changes or concavity reasoning.
