AP Calc BC Unit 3 Review: Chain Rule, Implicit Diff & More
AP Calc BC Unit 3 covers differentiation rules and their applications: the chain rule for composite functions, implicit differentiation for relations that aren’t solved for y, derivatives of inverse functions (including inverse trig), and related rates problems that connect these tools to real-world changing quantities.
An ap calc bc unit 3 review is really about recognizing which differentiation tool a problem demands. The BC course includes everything in AB plus additional topics, so Unit 3 content is the same core material tested on both exams—but BC problems often nest these rules inside later contexts like parametric or polar differentiation.
Unit 3 Topics Overview
Unit 3 is the differentiation-rules unit. It maps onto four connected skills.
- Chain rule: differentiating composite functions like sin(3x²) or e^(x³).
- Implicit differentiation: finding dy/dx when a relation like x² + y² = 25 isn’t solved for y.
- Inverse functions: derivatives of inverse trig functions and the general inverse-function derivative formula.
- Related rates: translating a “rates of change” word problem into an equation, then differentiating with respect to time.
The connective tissue is composition. The chain rule is the engine behind implicit differentiation—when you differentiate y with respect to x, you’re really applying the chain rule and multiplying by dy/dx. Inverse function derivatives also lean on the chain rule. Related rates are the applied payoff: every related rates problem is a chain rule problem where the variable you differentiate with respect to is time.
Because AP Calculus BC includes all AB material plus integration and series topics, Unit 3 is foundational for everything that follows. If chain rule or implicit differentiation is shaky here, later units—especially differential equations and series—become much harder.
Chain Rule Essentials
The chain rule states that for a composite function f(g(x)), the derivative is f′(g(x)) · g′(x). In plain terms: differentiate the outside function, keep the inside untouched, then multiply by the derivative of the inside.
Common AP-style forms worth memorizing:
- Power rule with chain: d/dx [u^n] = n·u^(n−1) · u′
- Exponential: d/dx [e^u] = e^u · u′
- Logarithm: d/dx [ln u] = u′ / u
- Trig: d/dx [sin u] = cos u · u′
Worked example
Differentiate y = (2x + 1)⁵.
- Outside: power of 5 → 5(2x + 1)⁴
- Inside: 2x + 1 → derivative 2
- Multiply: y′ = 10(2x + 1)⁴
A frequent AP trap is forgetting the inside derivative entirely—leaving out the · u′ factor. Slow down and name your inside function before differentiating.
Implicit Differentiation Steps
Use implicit differentiation when y is not isolated. Any equation where x and y are mixed—like x² + y² = 25—qualifies.
The procedure:
- Differentiate both sides with respect to x.
- Every time you differentiate a term containing y, multiply by dy/dx (this is the chain rule in action).
- Collect all dy/dx terms on one side.
- Factor out dy/dx and solve.
Worked example
Find dy/dx for x² + y² = 25.
- Differentiate: 2x + 2y · (dy/dx) = 0
- Isolate: 2y · (dy/dx) = −2x
- Solve: dy/dx = −x/y
Notice this gives the slope of the tangent line at any point on the circle, not just the top or bottom half. That’s the power of implicit differentiation: it handles relations that aren’t single-valued functions.
Derivatives of Inverse Functions
The general inverse-function derivative rule: if g is the inverse of f, then g′(x) = 1 / f′(g(x)), provided f′(g(x)) ≠ 0. Geometrically, slopes of inverse functions are reciprocals at corresponding points.
For inverse trig functions, memorize these:
| Function | Derivative |
|---|---|
| arcsin u | u′ / √(1 − u²) |
| arccos u | −u′ / √(1 − u²) |
| arctan u | u′ / (1 + u²) |
| arcsec u | u′ / (|u|√(u² − 1)) |
The arccos derivative is just the negative of arcsin, and arctan is the cleanest of the set. Apply the chain rule by treating u as your inside function.
Related Rates
Related rates problems use implicit differentiation with respect to time. A classic: a spherical balloon inflates at 100 cm³/s; how fast is the radius growing when r = 5 cm?
- Volume: V = (4/3)πr³
- Differentiate with respect to t: dV/dt = 4πr² · dr/dt
- Substitute: 100 = 4π(25) · dr/dt
- Solve: dr/dt = 1/π ≈ 0.318 cm/s
The setup is identical to implicit differentiation—only the variable you differentiate against changes. Label every rate as d(something)/dt before you start, and always identify what’s given versus what’s asked.
For extra practice, Khan Academy has targeted chain rule and implicit differentiation drills, and the official College Board course description lists the exact learning objectives for this unit. The broader AP Calculus structure is useful context if you’re planning your full-year study sequence.
Key Takeaways
- Chain rule: f′(g(x)) · g′(x) — never skip the inside derivative.
- Implicit differentiation: differentiate both sides, attach dy/dx to every y-term, then solve.
- Inverse functions: g′(x) = 1 / f′(g(x)); memorize the inverse trig derivatives.
- Related rates: implicit differentiation with respect to time — identify given and unknown rates first.
- Unit 3 skills reappear throughout BC in differential equations, series, and parametric problems, so fluency here pays off repeatedly.
