AP Syllabus focus:
‘A critical point on an implicit relation occurs where dy/dx equals zero or does not exist, provided the point lies on the curve defined by the relation.’
Implicit relations arise when an equation combines x and y without solving explicitly for one variable, so identifying critical points requires careful reasoning about the derivative’s behavior along the curve.
Understanding Critical Points on Implicit Relations
A critical point is a location on a curve where its instantaneous rate of change meets specific conditions. For implicit relations, these points occur where the derivative with respect to x is either zero or undefined. Because the curve is not written as y = f(x), we rely on implicit differentiation to uncover the behavior of the relation and determine where these special points occur.
Critical Point: A point on a curve where the derivative equals zero or is undefined, assuming the point lies within the domain of the original relation.
A valid critical point must satisfy two conditions: it must lie on the curve itself, and the slope information from the derivative must reveal either a horizontal or vertical tangent. These conditions ensure that the point genuinely reflects meaningful behavior in the geometric structure of the relation.

This graph shows a smooth curve with a tangent line touching it at one point, illustrating how the slope of the tangent represents . It demonstrates how each critical point on an implicit relation is detected by evaluating the slope of its tangent line. Source.
Interpreting dy/dx for Implicit Relations
To determine critical points, we first compute by differentiating the implicit equation with respect to x, treating y as a differentiable function. This process uses familiar rules such as the product rule and chain rule, but applied within the more general setting of an equation involving both variables. The resulting derivative often appears as a rational expression involving x, y, and sometimes multiple combined terms.
= Slope of the tangent line to the implicit curve at a point
After determining this expression, we analyze where the derivative equals zero or fails to exist. These locations correspond to potential horizontal or vertical tangent lines, respectively. Because tangent behavior reflects local structure, such points mark where the curve changes direction, levels out, or becomes momentarily vertical in its orientation.
A key distinction arises when interpreting these cases. A derivative equal to zero indicates that the curve momentarily has no rise relative to run, forming a horizontal tangent. A derivative failing to exist suggests the slope becomes infinite or undefined, which corresponds to a vertical tangent. Both provide valuable insight into how the curve behaves in a neighborhood of the point.

This image shows a curve with a clearly marked vertical tangent at the origin, where the derivative becomes undefined. It visually demonstrates how a vertical tangent corresponds to a point where does not exist, a key condition for critical points on implicit relations. Source.
Conditions Required to Identify a Critical Point
A point where or is undefined does not automatically qualify as a critical point. The point must also satisfy the original implicit equation. This requirement preserves the logical connection between the derivative and the curve itself: slope values only make sense along points that belong to the relation. Therefore, verifying that each candidate point lies on the curve is essential.
To ensure accuracy when identifying critical points on implicit curves, students should follow a clear, logical process:
Differentiate the relation implicitly, obtaining an expression for that reflects how both variables contribute to the slope.
Determine where the derivative equals zero, indicating potential horizontal tangents.
Find where the derivative is undefined, revealing points with potential vertical tangents.
Check whether each candidate satisfies the original implicit equation, guaranteeing the point lies on the curve.
Confirm that the x-value and y-value are within the domain of the relation, ensuring the point is meaningful within the context of the curve.
These steps help differentiate between algebraic possibilities and true geometric features of the implicit curve.
Geometric Meaning of Critical Points on Implicit Relations
When , the slope of the tangent line is horizontal. Geometrically, this corresponds to locations where the curve peaks, bottoms out, or otherwise levels before changing direction. Although identifying the precise nature of this behavior requires additional analysis, locating the point itself relies on the derivative structure.
When does not exist, the slope is vertical. Geometrically, the curve rises or falls without any horizontal motion, such as in a sideways cusp or a vertical tangent segment. These situations arise naturally in implicit relations, especially when the equation cannot be expressed cleanly as y as a function of x.
In both cases, the point’s validity as a critical point hinges on whether it belongs to the curve defined by the relation. The derivative alone cannot determine membership; only substitution into the original equation confirms the point’s presence on the curve.

This diagram presents the implicit curve with a tangent line at the point , demonstrating how implicit differentiation identifies tangent behavior. It reinforces that critical points must satisfy both the derivative condition and the original implicit equation. Source.
Why Critical Points Matter for Implicit Curves
Critical points help illuminate a curve’s structure by identifying where important geometric behaviors occur. On implicit curves, these points often reveal turning behavior, cusp-like features, or changes in orientation. Understanding how derivatives behave in this context deepens students’ comprehension of calculus as a tool for analyzing curves beyond explicit function forms.
FAQ
A horizontal tangent on an implicit curve occurs when dy/dx equals zero at a point that lies on the implicit relation, but this does not necessarily imply a maximum or minimum because the curve might bend or turn differently from an explicit function.
An explicit stationary point requires y to be a defined function of x locally, whereas an implicit curve may not behave like a single-valued function near the point. This means a horizontal tangent on an implicit curve might correspond to a turning point, a cusp-like feature, or a point where multiple branches meet.
Yes. Some implicit relations produce cusp points or sharp corners where the curve has no well-defined tangent direction. In such cases, dy/dx may not exist because the slope approaches different values from different directions.
A point like this can still be considered a critical point if it lies on the curve and the derivative is undefined. The absence of a tangent is itself a geometric feature relevant to the curve’s behaviour.
To check this, examine the derivative expression near the point rather than only at the point.
Steps include:
• Evaluate the numerator and denominator behaviour around the point.
• Determine whether the denominator approaches zero smoothly while the numerator stays non-zero.
• Confirm the graph locally resembles a continuous curve with a vertical direction rather than a broken or jump-like feature.
If the curve remains continuous and the slope becomes unbounded, the tangent is genuinely vertical.
This can happen only in exceptional cases where the curve self-intersects or has multiple branches passing through the same coordinate. Different branches may have different tangent behaviours at the same point.
For example, one branch might have dy/dx equal to zero, while another branch has dy/dx undefined. Identifying these situations requires checking all possible solutions for dy/dx that arise from the implicit equation.
Implicit differentiation produces conditions on x and y that might not satisfy the original relation. A coordinate pair that solves dy/dx equals zero or makes dy/dx undefined is only meaningful if it also satisfies the equation defining the curve.
Checking the point ensures that the geometric interpretation of the derivative matches an actual location on the curve and prevents false positives arising from algebraic manipulation of the derivative expression.
Practice Questions
The curve is defined implicitly by the equation
x squared plus y squared minus 4y equals 0.
(a) Find the value of dy/dx at any point on the curve.
(b) Hence determine the x-coordinate of any point on the curve where the tangent is horizontal.
(1–3 marks)
Question 1
(a) One mark for correct differentiation of each term and correct expression for dy/dx.
Award 1 mark for obtaining:
dy/dx = (4y − 2x) divided by (2y).
(b) One mark for setting dy/dx equal to zero.
One mark for correctly solving for x using 4y − 2x = 0 to obtain x = 2y.
(If linked with the original curve to find an x-value consistent with the curve, award full marks.)
Total: 3 marks
The curve C is defined by the implicit relation
x cubed plus y cubed equals 6xy.
(a) Use implicit differentiation to find an expression for dy/dx.
(b) Show that the point (2,2) lies on C and determine whether the tangent at this point is horizontal, vertical, or neither.
(c) Verify that your candidate in part (b) satisfies the condition for a critical point of an implicit relation.
(4–6 marks)
Question 2
(a) One mark for correctly applying implicit differentiation to x cubed and y cubed.
One mark for correctly differentiating 6xy using the product rule.
One mark for obtaining a correct simplified expression for dy/dx.
Accept:
dy/dx = (6y − 3x squared) divided by (3y squared − 6x).
(b) One mark for showing substitution of (2,2) into the original relation: 8 + 8 = 24.
One mark for identifying this as true, so (2,2) lies on C.
One mark for substituting into dy/dx to determine tangent behaviour.
Award full credit for concluding that the tangent is neither horizontal nor vertical because dy/dx is non-zero and finite.
(c) One mark for verifying the definition of a critical point: dy/dx equals zero or does not exist, and the point lies on the curve.
One mark for concluding correctly that (2,2) is not a critical point because dy/dx is neither zero nor undefined.
Total: 6 marks
