Differentiation: Composite, Implicit & Inverse Functions
What this unit covers
The topics below follow the published Calculus BC course framework for Unit 3. This unit is worth 5–10% of the exam, so budget your time against that rather than against how long the unit takes to teach.
Lessons in this unit
- The Chain Rule14 min · 3 objectivesDifferentiate compositions of functions with the chain rule · Identify the outer and inner functions in a composition · Combine the chain rule with the product and quotient rules
- Implicit Differentiation14 min · 3 objectivesDifferentiate equations that are not solved for y using implicit differentiation · Apply the chain rule to y-terms, attaching dy/dx · Solve for dy/dx and evaluate the slope at a given point
- Derivatives of Inverse Functions13 min · 3 objectivesApply the inverse-function derivative formula · Differentiate inverse trigonometric functions · Relate the slopes of a function and its inverse at corresponding points
- Nested Chains and Higher-Order Derivatives15 min · 3 objectivesDifferentiate a composition of three or more functions by applying the chain rule repeatedly · Compute a second derivative implicitly and substitute dy/dx before simplifying · Combine the chain rule with the product and quotient rules in a single expression
Formulas in Unit 3
Every term in Unit 3
All 22 terms we publish for Differentiation: Composite, Implicit & Inverse Functions, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.
- Chain rule
- d/dx[f(g(x))] = f′(g(x))·g′(x). Differentiate the outside leaving the inside alone, then multiply by the inside's derivative.
- Nested chain rule
- For f(g(h(x))) the factors multiply outward: f′(g(h))·g′(h)·h′(x). Missing the innermost factor is the usual slip.
- Implicit differentiation
- Differentiate both sides with respect to x, applying the chain rule to every y term so each yields dy/dx, then solve.
- Second derivative implicitly
- Differentiate dy/dx again, then substitute the expression for dy/dx wherever it reappears.
- Vertical and horizontal tangents implicitly
- Horizontal where the numerator of dy/dx is zero, vertical where the denominator is — and the point must lie on the curve.
- Derivative of an inverse trig composite
- Combine the inverse trig derivative with the chain rule: d/dx[arctan(u)] = u′/(1 + u²).
- Differentiating a parametrically defined function
- dy/dx = (dy/dt)/(dx/dt), which is the chain rule rearranged rather than a separate rule.
- Implicit differentiation is the chain rule on y
- Every y term picks up dy/dx, exactly as any inner function would. d/dx[y³] = 3y²(dy/dx).
- The xy term needs the product rule
- d/dx[xy] = y + x(dy/dx) — two terms. Forgetting the second is the single most common implicit error.
- Why dy/dx contains y
- An implicit relation need not be a function, so the slope depends on the branch. That is why such questions supply a point, not an x value.
- Horizontal tangent, implicit
- Numerator of dy/dx zero, denominator nonzero — and the point must also satisfy the curve equation.
- Vertical tangent, implicit
- Denominator zero, numerator nonzero. If both vanish, the point is indeterminate and needs separate analysis.
- Inverse derivative evaluation point
- g′(b) = 1/f′(a) where f(a) = b. Evaluate f′ at a, the ORIGINAL input — using b is the standard error.
- Finding a for an inverse problem
- The exam almost always makes a a small integer. Test 0, ±1, ±2 before attempting algebra.
- Work a nested chain outside in
- Name the layers first. For sin³(4x² + 1) they are cube, sine, quadratic, and each contributes a factor.
- The lost inner factor
- Multi-layer chain-rule errors are almost always the missing derivative of the innermost expression, because it is written last.
- Chain rule with a table
- h′(a) = f′(g(a))·g′(a). The outer derivative is evaluated at the OUTPUT of g — that substitution is what the rule is about.
- Derivative of ln u
- u′/u. The inner derivative goes in the numerator, which is where sign errors appear.
- Derivative of e^u
- e^u·u′. The exponential reproduces itself, so the only work is the inner derivative.
- Derivative of arcsin
- 1/√(1 − u²) times u′, defined only for |u| < 1 — the domain restriction is part of the answer.
- Higher-order derivatives of sine
- The pattern sin, cos, −sin, −cos repeats every four derivatives, so the nth derivative depends on n mod 4 alone.
- Composition of three functions
- f(g(h(x)))′ = f′(g(h(x)))·g′(h(x))·h′(x). Each factor is evaluated at everything inside it.
What examiners penalize here
- Chain rule questions escalate by nesting. For sin³(2x) = (sin(2x))³ you apply the rule twice: 3(sin 2x)² · cos(2x) · 2. Count the layers before you start so you know how many inner-derivative factors to expect.
- Implicit results usually contain both x and y — that is correct, not incomplete. To report a numerical slope you must plug in a full point (x, y) on the curve. Free-response graders expect the point substituted, not just the symbolic derivative.
- Inverse-function problems on the AP exam almost always give you a table or a point pairing. The move is mechanical: find where f equals the target output, evaluate f′ there, take the reciprocal. Practice it until it is automatic.
- Nested chain rule questions are frequently multiple choice with all the right pieces and one missing factor among the distractors. Count your factors against the number of layers before choosing.
Practice Calculus BC
Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.
Questions about this unit
How much of the AP Calculus BC exam is Unit 3?
Unit 3, Differentiation: Composite, Implicit & Inverse Functions, is worth 5–10% of the Calculus BC multiple-choice section according to the published course framework. Across all 10 units that makes it a middling share, roughly what an even split across units would give.
What topics are covered in Calculus BC Unit 3?
Differentiation: Composite, Implicit & Inverse Functions covers Chain rule, Implicit, Inverse functions and Higher-order. We publish 22 terms with definitions for this unit, all of them on this page.
How should I study Calculus BC Unit 3?
Read the 4 lessons below first — about 55 minutes — then drill the 22 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.
All 10 units of AP Calculus BC
- Unit 1 · Limits & Continuity
- Unit 2 · Differentiation: Definition & Fundamental Properties
- Unit 3 · Differentiation: Composite, Implicit & Inverse Functions
- Unit 4 · Contextual Applications of Differentiation
- Unit 5 · Analytical Applications of Differentiation
- Unit 6 · Integration & Accumulation of Change
- Unit 7 · Differential Equations
- Unit 8 · Applications of Integration
- Unit 9 · Parametric, Polar & Vector-Valued Functions
- Unit 10 · Infinite Sequences & Series
Unit names, topics and exam weights follow the published College Board course framework for AP Calculus BC. AP® is a trademark registered by the College Board, which does not endorse this site.