Composite & Implicit Differentiation

Name: AP Calculus AB Unit 3: Composite & Implicit Differentiation
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5–10% of the exam6 lessons · 84 min38 terms

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What this unit covers

The topics below follow the published Calculus AB 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.

ImplicitInverse functionsHigher-orderRelated contexts

Lessons in this unit

Formulas in Unit 3

Chain rule on y-terms

d/dx[yⁿ] = n·yⁿ⁻¹ · (dy/dx)

Treat y as a function of x. Any term containing y produces a dy/dx factor; terms in x alone do not.

Derivative of an inverse function

If g = f⁻¹, then g′(a) = 1 / f′(g(a))

The inverse’s slope at a is the reciprocal of f’s slope at the matching point b = g(a) where f(b) = a.

Key inverse trig derivatives

d/dx[arcsin x] = 1/√(1 − x²) · d/dx[arctan x] = 1/(1 + x²)

And d/dx[arccos x] = −1/√(1 − x²). Chain rule applies for arcsin(u): u′/√(1 − u²).

Second derivative and motion

v(t) = s′(t) · a(t) = s″(t) = v′(t)

Velocity is the first derivative of position; acceleration is the second derivative of position (the derivative of velocity).

Exponential and log derivatives

d/dx[eˣ] = eˣ · d/dx[ln x] = 1/x · d/dx[aˣ] = aˣ·ln a

With the chain rule: d/dx[e^u] = e^u·u′ and d/dx[ln u] = u′/u.

Implicit differentiation, worked shape

From x² + y² = 25: 2x + 2y·(dy/dx) = 0 ⟹ dy/dx = −x/y

Every y term contributes dy/dx by the chain rule. Collect those terms, factor, and divide.

Inverse function derivative

If g = f⁻¹, then g′(b) = 1 / f′(a) where f(a) = b

Reciprocal of the derivative of f, evaluated at the ORIGINAL point a, not at b. Find a first.

Every term in Unit 3

All 38 terms we publish for Composite & Implicit Differentiation, 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 derivative of the inside.
Implicit differentiation: Differentiate both sides with respect to x, applying the chain rule to every y term so each contributes dy/dx, then solve for dy/dx.
Recognizing a composite: Anything raised to a power, or a trig, exponential or log function of something other than plain x, needs the chain rule.
Vertical and horizontal tangents implicitly: Horizontal where the numerator of dy/dx is zero, vertical where the denominator is zero — provided the point is on the curve.
Derivative of an inverse function: (f⁻¹)′(b) = 1/f′(a) where f(a) = b. Find the input a first; the formula uses the original function's derivative there.
Derivatives of inverse trig functions: arcsin′ = 1/√(1 − x²), arccos′ = −1/√(1 − x²), arctan′ = 1/(1 + x²).
Logarithmic differentiation: Take ln of both sides before differentiating. Handles variable bases with variable exponents and long products.
Second derivative implicitly: Differentiate dy/dx again with respect to x, then substitute the expression for dy/dx wherever it appears.
Implicit differentiation is the chain rule on y: Because y depends on x, d/dx[y³] = 3y²(dy/dx). Every y term picks up dy/dx, exactly as any inner function would.
Collect and factor: After differentiating, gather every dy/dx term on one side, factor it out, and divide. The result generally involves both x and y.
Why dy/dx involves y: An implicit relation need not be a function, so the slope depends on which branch you are on. That is why questions give a point rather than an x value.
Horizontal tangent from an implicit derivative: Set the NUMERATOR of dy/dx to zero, and require the denominator to be nonzero there.
Vertical tangent from an implicit derivative: Set the DENOMINATOR to zero, and require the numerator to be nonzero there.
Candidate points must lie on the curve: Solve the numerator or denominator condition simultaneously with the original equation. A point satisfying only one is not on the curve.
Numerator and denominator both zero: Neither the horizontal nor the vertical conclusion follows; the point needs separate analysis and often signals a singular point.
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-differentiation error.
Circle slope formula: For x² + y² = r², dy/dx = −x/y. Vertical tangents where y = 0, horizontal where x = 0 — the geometry confirms both.
Sign of an implicit second derivative: Once d²y/dx² is expressed in x and y, evaluate it at a point on the curve: a positive value means concave up there, exactly as for an explicit function.
Work outside in: Name the layers before differentiating. For sin³(4x² + 1) the layers are cube, sine, quadratic, and each contributes a factor.
The lost inner factor: Multi-layer chain rule errors are almost always a missing derivative of the innermost expression, because it is written last.
Chain inside product: Decide which rule governs the OUTERMOST structure first. x²sin(3x) is a product containing a composition; sin(x²eˣ) is a composition containing a product.
Inverse derivative formula: g′(b) = 1/f′(a) where f(a) = b. Reciprocal because reflecting across y = x swaps rise and run.
Where to evaluate an inverse derivative: At a, the original input, not at b. Find a with f(a) = b first — the exam nearly always makes a a small integer.
Derivative of arctan: d/dx[arctan x] = 1/(1 + x²). Always positive, consistent with arctangent increasing everywhere.
Derivative of arcsin: d/dx[arcsin x] = 1/√(1 − x²), defined only on (−1, 1) — the domain restriction is part of the answer.
Chain rule with an exponential: d/dx[e^u] = e^u·u′. The exponential reproduces itself, so the only work is the inner derivative.
Chain rule with a logarithm: d/dx[ln u] = u′/u. Note the inner derivative goes in the numerator, which is where sign errors appear.
Derivative of a variable base with a variable exponent: For xˣ neither the power rule nor the exponential rule applies. Rewrite as e^(x ln x), or take ln of both sides and differentiate implicitly.
Derivative of a variable base and exponent: For xˣ, neither the power rule nor the exponential rule applies. Rewrite as e^(x ln x) or use logarithmic differentiation.
Higher-order derivatives of sine: The pattern sin, cos, −sin, −cos repeats every four derivatives, so the nth derivative depends only on n mod 4.
Chain rule with a table: h′(a) = f′(g(a))·g′(a). The outer derivative is evaluated at the OUTPUT of g, which is the substitution the rule is really about.
Related rates as implicit differentiation: Differentiating a geometric relation with respect to t is implicit differentiation with t as the variable — the same mechanism, a different letter.
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.
Implicit differentiation with trig: d/dx[sin y] = cos y·(dy/dx). The chain-rule factor is easy to lose because sine and cosine look like a closed operation.
Tangent line to an implicit curve: Compute dy/dx, verify the point lies on the curve, substitute BOTH coordinates, then use point-slope form.
Two slopes at one x: On x² + y² = 25 at x = 3 the slopes are −3/4 and +3/4 at (3, 4) and (3, −4). Not an error — the point of the method.
Implicit second derivative sign: Once d²y/dx² is expressed in x and y, its sign at a point gives concavity there, exactly as for an explicit function.
Inverse trig chain rule: d/dx[arctan(u)] = u′/(1 + u²). The inner derivative goes in the numerator, matching the logarithm pattern.

What examiners penalize here

To find a tangent slope on an implicit curve you need both coordinates, because dy/dx typically depends on x *and* y. Differentiate, solve for dy/dx, then substitute the full point — not just the x-value.
The formula g′(a) = 1/f′(g(a)) shows up on the AP exam most often in table form: you are handed values of f and f′ and asked for the inverse’s derivative. Identify the matching point from the table, then take the reciprocal.
On motion problems, read the question’s verb carefully: "how fast" usually means velocity (first derivative), while "how is the velocity changing" means acceleration (second derivative). Matching the derivative order to the phrasing is worth easy points.
When several rules stack, name the outermost operation before computing anything. "This is a product, whose second factor is a composite" gives you a plan and keeps the chain-rule factors from getting lost in a long expression.

Practice Calculus AB

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.

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Questions about this unit

How much of the AP Calculus AB exam is Unit 3?

Unit 3, Composite & Implicit Differentiation, is worth 5–10% of the Calculus AB multiple-choice section according to the published course framework. Across all 8 units that makes it a middling share, roughly what an even split across units would give.

What topics are covered in Calculus AB Unit 3?

Composite & Implicit Differentiation covers Implicit, Inverse functions, Higher-order and Related contexts. We publish 38 terms with definitions for this unit, all of them on this page.

How should I study Calculus AB Unit 3?

Read the 6 lessons below first — about 85 minutes — then drill the 38 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 8 units of AP Calculus AB

Unit names, topics and exam weights follow the published College Board course framework for AP Calculus AB. AP® is a trademark registered by the College Board, which does not endorse this site.