An implicit relation is an equation connecting x and y that is not solved as y = f(x), like x² + y² = 25. In AP Calculus, you analyze these curves with implicit differentiation, finding critical points where dy/dx equals zero or does not exist (Topic 5.12).
An implicit relation is an equation where x and y are tangled together instead of cleanly separated. There's no "y =" on one side. The circle x² + y² = 25 is the classic example. You can't write that whole circle as a single function y = f(x), because most x-values pair with two y-values, and a function only allows one output per input.
Even though you can't isolate y, calculus still works. You differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule every time you hit a y. The result is dy/dx, usually written as an expression involving both x and y. From there, everything you learned about analyzing functions still applies. A critical point is a point on the curve where dy/dx equals zero or doesn't exist, exactly as the CED defines it for Topic 5.12. Second derivatives get messier, since d²y/dx² may involve x, y, and dy/dx all at once, but the logic is the same.
Implicit relations live in Topic 5.12 (Exploring Behaviors of Implicit Relations) in Unit 5: Analytical Applications of Differentiation. Two learning objectives drive this topic. AP Calc 5.12.A asks you to determine critical points of implicit relations, and AP Calc 5.12.B asks you to justify conclusions about the behavior of an implicitly defined function using evidence from its derivatives.
Here's the big idea behind the topic. Everything from Unit 5, like finding where a curve has horizontal tangents, where it increases or decreases, and where it's concave up or down, extends to curves that aren't functions at all. The exam loves this because it tests whether you actually understand what dy/dx means, or whether you just memorized procedures for y = f(x). If dy/dx is a fraction, a horizontal tangent happens where the numerator is zero (and the denominator isn't), and a vertical tangent happens where the denominator is zero. Being able to reason that way, with both x and y in play, is the whole point.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryImplicit Differentiation (Unit 3)
Topic 3.2 teaches the technique; Topic 5.12 applies it. You differentiate both sides with respect to x, using the chain rule on every y term, and that dy/dx expression is the raw material for every critical point and concavity question in 5.12.
Explicit Relation (Unit 1+)
An explicit relation hands you y = f(x) directly. An implicit relation is the same idea with the algebra left unsolved. Most of AP Calc runs on explicit functions, so 5.12 is the moment the course proves those tools work even when you can't isolate y.
Critical Points and Curve Analysis (Unit 5)
Topics 5.2 through 5.7 build the toolkit of critical points, increasing/decreasing intervals, and concavity for regular functions. Topic 5.12 is the same toolkit pointed at implicit curves, with one twist. Since dy/dx contains both x and y, you usually need to plug points back into the original equation to finish.
Parametric Equations (Unit 9, BC only)
Parametric curves are another way to describe shapes that fail the vertical line test, like circles. If you're taking BC, the dy/dx = (dy/dt)/(dx/dt) logic in Unit 9 will feel familiar, because it's the same game of finding slopes on a curve that isn't y = f(x).
Implicit relations show up in multiple-choice questions that test both the setup and the interpretation. Common stems ask you to identify what counts as an implicit relation, compute dy/dx by differentiating both sides with respect to x, or interpret what dy/dx = 0 means at a point on the curve (a horizontal tangent line).
On free-response questions, implicit relations typically appear as a multi-part analysis problem. A classic structure gives you a curve like x² + xy + y² = k and asks you to (a) find dy/dx, (b) find all points with a horizontal or vertical tangent, and (c) use the second derivative to justify concavity or classify a point. The justification language matters. Saying "dy/dx = 0 and the point lies on the curve" earns points; just solving the numerator equals zero without checking the point is actually on the relation does not. Expect the second derivative to involve x, y, and dy/dx together, and expect to substitute known values rather than simplify everything symbolically.
An explicit relation solves for y, like y = √(25 − x²). An implicit relation leaves x and y mixed, like x² + y² = 25. The implicit version often describes a bigger curve (the full circle, not just the top half), and its derivative dy/dx usually depends on both x and y. With an explicit function, dy/dx depends on x alone.
An implicit relation is an equation relating x and y that is not written in the form y = f(x), such as x² + y² = 25.
To find dy/dx for an implicit relation, differentiate both sides of the equation with respect to x and apply the chain rule to every term containing y.
A critical point of an implicit relation is a point on the curve where dy/dx equals zero or does not exist, per the essential knowledge for Topic 5.12.
When dy/dx is a fraction, horizontal tangents occur where the numerator is zero, and vertical tangents occur where the denominator is zero, as long as the point actually satisfies the original equation.
Second derivatives of implicit relations can contain x, y, and dy/dx, so substitute your known dy/dx expression and point values instead of trying to fully simplify.
Always verify candidate points lie on the original relation; solving dy/dx = 0 alone is not a complete justification on the FRQ.
It's an equation where x and y are related but y is not isolated, like x² + y² = 25 or x³ + y³ = 6xy. You analyze these curves in Topic 5.12 using implicit differentiation to find dy/dx and study the curve's behavior.
An explicit function gives you y = f(x) directly, so dy/dx depends only on x. An implicit relation leaves x and y mixed together, so dy/dx typically depends on both variables, and the curve may fail the vertical line test entirely.
No, and that's the point. The circle x² + y² = 25 fails the vertical line test, so it's not a function. But near most points, a piece of the curve behaves like a function, which is why calculus tools like critical points still apply.
It means the curve has a horizontal tangent line at that point. This is exactly how Fiveable and AP practice questions test the concept, and it's the first step in finding critical points under learning objective 5.12.A.
First find dy/dx by implicit differentiation. Then find points where dy/dx equals zero (numerator zero) or does not exist (denominator zero), and confirm each point satisfies the original equation. Points not on the curve don't count.