An implicitly defined function is a function described by an equation relating x and y (like x³ + y³ = 6xy) instead of an explicit formula y = f(x). On the AP exam, you differentiate it with implicit differentiation (Topic 3.2) and analyze its behavior using dy/dx and d²y/dx² (Topic 5.12).
An implicitly defined function is a function hiding inside an equation. Instead of handing you y = f(x) directly, the equation tangles x and y together, like x² + y² = 25 or x³ + y³ = 6xy. The curve still has well-defined slopes and behavior at most points, you just can't (or don't want to) solve for y first.
Think of it this way. An explicit function tells you exactly how to compute y from x. An implicit equation just tells you the rule that x and y must satisfy together. Calculus still works on these curves because the chain rule lets you differentiate y-terms without ever isolating y. That's the entire idea behind implicit differentiation: treat y as a function of x, so differentiating y³ gives 3y² · dy/dx. Your answer for dy/dx will usually involve both x and y, and that's expected.
This term anchors two CED learning objectives. In Unit 3 (Topic 3.2), LO 3.2.A asks you to calculate derivatives of implicitly defined functions, with the essential knowledge that the chain rule is the basis for implicit differentiation. In Unit 5 (Topic 5.12), LO 5.12.A and LO 5.12.B push it further: find critical points of implicit relations (where dy/dx equals zero or does not exist) and justify conclusions about behavior using derivative evidence. The CED also flags that second derivatives of implicit functions may be relations of x, y, AND dy/dx, which is a classic place to lose points. Beyond these two topics, the skill quietly powers related rates and derivatives of inverse functions, so it shows up on the exam far more often than the name does.
Keep studying AP® Calculus Unit 5
Visual cheatsheet
view galleryImplicit relation (Units 3 and 5)
An implicit relation is the full equation, like x² + y² = 25, while an implicitly defined function is a piece of that curve that actually passes the vertical line test (the top or bottom half of the circle, for instance). The CED uses both terms because one relation can contain several functions.
The chain rule (Unit 3)
Implicit differentiation IS the chain rule in disguise. Since y is secretly a function of x, every derivative of a y-term picks up a dy/dx factor. If you understand why d/dx(y²) = 2y · dy/dx, you understand the whole technique.
Related rates (Unit 4)
Related rates problems are implicit differentiation with respect to time instead of x. Every variable in the equation gets a dt-derivative attached, which is the same chain-rule move you learned for implicitly defined functions.
Critical points and curve analysis (Unit 5)
Everything you do with explicit functions in Unit 5 (critical points, increasing/decreasing, concavity) extends to implicit curves. The twist is that dy/dx depends on both x and y, so you plug in a full point, not just an x-value, and the second derivative may contain dy/dx itself.
Multiple-choice questions usually hand you an equation like x³ + y³ = 6xy and a point, then ask for the slope, the tangent line, or a behavior claim (increasing, concave down, has a horizontal tangent) at that point. Practice questions also test the concept itself, like asking what advantage implicit differentiation gives you (you never have to solve for y). On free-response, this skill appears inside larger problems: finding where a curve has horizontal or vertical tangents, justifying behavior with dy/dx, or computing d²y/dx² and substituting your dy/dx expression back in. The non-negotiable mechanics are attaching dy/dx to every differentiated y-term, evaluating at the full (x, y) point, and remembering that dy/dx = 0 means horizontal tangent while an undefined dy/dx (denominator zero) often means a vertical tangent.
An explicit function gives you a recipe, y = f(x), so you can compute y directly from x. An implicitly defined function is buried in an equation relating x and y, and solving for y may be ugly or impossible. The calculus payoff is the same (slopes, tangents, concavity), but the method changes. Explicit functions use ordinary differentiation, while implicit ones require the chain rule on every y-term, producing a dy/dx that typically involves both x and y.
An implicitly defined function comes from an equation relating x and y, like x³ + y³ = 6xy, rather than an explicit formula y = f(x).
The chain rule is the engine of implicit differentiation, so every y-term you differentiate must carry a dy/dx factor (this is LO 3.2.A).
Your dy/dx for an implicit function usually contains both x and y, so you evaluate it at a complete point, not just an x-value.
A critical point of an implicit relation is where dy/dx equals zero or does not exist (LO 5.12.A), and a zero numerator with nonzero denominator means a horizontal tangent.
Second derivatives of implicitly defined functions can include x, y, and dy/dx, so substitute your dy/dx expression back in before evaluating.
All the Unit 5 analysis tools, including increasing/decreasing behavior and concavity, extend to implicitly defined functions (LO 5.12.B).
It's a function described by an equation that mixes x and y together, like x² + y² = 25, instead of giving you y = f(x) directly. You differentiate it using implicit differentiation, which is Topic 3.2 in the AP Calc CED.
No, and that's the whole point. Implicit differentiation lets you find dy/dx without ever isolating y, which matters because many implicit equations (like x³ + y³ = 6xy) can't be cleanly solved for y at all.
The implicit relation is the entire equation and its full curve, which may fail the vertical line test. An implicitly defined function is a portion of that curve that does behave as a function, like the top half of the circle x² + y² = 25.
Because the chain rule attaches dy/dx to every y-term, and when you solve for dy/dx, leftover y's stay in the formula. That's why exam questions give you a full point like (2, 1) to plug in, not just an x-value.
Find where dy/dx equals zero or does not exist, per LO 5.12.A. Typically dy/dx is a fraction, so set the numerator equal to zero for horizontal tangents and check where the denominator is zero for points where the derivative doesn't exist.
Connect this key term to the AP exam workflow: review the course, practice questions, and check related study tools.
Review units, study guides, and course resources.
Check this vocabulary in multiple-choice context.
Apply key concepts in written AP responses.
Estimate the exam score you are working toward.
Review the highest-yield facts before practice.