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 4–7% 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

Formulas in Unit 3

Every term in Unit 3

All 7 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.

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.

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

All 10 units of AP Calculus BC

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Fundamental Properties
3. Unit 3 · Differentiation: Composite, Implicit & Inverse Functions
4. Unit 4 · Contextual Applications of Differentiation
5. Unit 5 · Analytical Applications of Differentiation
6. Unit 6 · Integration & Accumulation of Change
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration
9. Unit 9 · Parametric, Polar & Vector-Valued Functions
10. Unit 10 · Infinite Sequences & Series