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Flashcards · AP® Calculus AB · Unit 3

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions flashcards

38 cards on topics 3.1 to 3.6. Pick a topic or study the whole unit.

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3.1 The Chain Rule

What is the chain rule?
If h(x) = f(g(x)), then h′(x) = f′(g(x))·g′(x): differentiate the outside, leave the inside alone, multiply by the derivative of the inside. Leibniz form: dy/dx = (dy/du)(du/dx).
How do you spot a hidden composite?
Ask "is the input exactly x?" If not, there is an inner function. sin²x, e^(4x), ln(3x) and 1/(x² + 1) all need the chain rule.
What are the chain-rule forms for powers, eˣ and ln?
d/dx [g(x)]ⁿ = n[g(x)]ⁿ⁻¹·g′(x); d/dx e^(g(x)) = e^(g(x))·g′(x); d/dx ln(g(x)) = g′(x)/g(x) for g(x) > 0; d/dx √g(x) = g′(x)/(2√g(x)).
What is the difference between the derivatives of sin²x and sin(x²)?
d/dx sin²x = 2 sin x cos x (the square is outside). d/dx sin(x²) = 2x cos(x²) (sine is outside).
g(2) = 7, g′(2) = 3 and f′(7) = −1. Find the derivative of f(g(x)) at x = 2.
f′(g(2))·g′(2) = f′(7)·3 = −3. Find g(a) first; using f′(2) is a common mistake.
Differentiate (4x − 1)⁵ and cos(x²).
(4x − 1)⁵ → 5(4x − 1)⁴·4 = 20(4x − 1)⁴. cos(x²) → −sin(x²)·2x = −2x sin(x²), not −sin(2x).
In Leibniz form, what units does a rate in metres per second times a rate in kelvin per metre give?
Kelvin per second: (dT/dx)(dx/dt) = dT/dt, and the units cancel like fractions.

3.2 Implicit Differentiation

What is the method for implicit differentiation?
Treat y as a function of x. Differentiate both sides, attaching dy/dx to every y term (chain rule); collect the dy/dx terms; factor out dy/dx; divide.
What is d/dx of y², xy and sin y?
d/dx y² = 2y·dy/dx; d/dx (xy) = y + x·dy/dx (product rule); d/dx sin y = cos y·dy/dx.
Find the slope of x² + y² = 10 at (1, 3).
2x + 2y·dy/dx = 0, so dy/dx = −x/y = −1/3.
Find the slope of xy = 6 at (2, 3).
y + x·dy/dx = 0, so dy/dx = −y/x = −3/2.
How do you find horizontal and vertical tangents from an implicit dy/dx = N/D?
Horizontal tangent: N = 0 with D ≠ 0. Vertical tangent: D = 0 with N ≠ 0. The points must also satisfy the original equation. If both are 0, the formula does not decide.
What mistakes are common in implicit differentiation?
Writing d/dx y² = 2y with no dy/dx; writing d/dx (xy) = dy/dx; differentiating the constant side to anything but 0; substituting only x into a dy/dx that contains y.

3.3 Differentiating Inverse Functions

What is the derivative of an inverse function?
If g is the inverse of f and f(a) = b, then g′(b) = 1/f′(a), provided f′(a) ≠ 0. In general g′(x) = 1/f′(g(x)).
Why does the inverse have reciprocal slopes?
The graph of f⁻¹ is f reflected in y = x: (a, b) moves to (b, a) and rise and run swap. Algebraically, differentiating f(g(x)) = x gives f′(g(x))·g′(x) = 1.
f(x) = x⁵ + 3x + 1 and g is its inverse. Find g′(5).
Solve f(a) = 5: a = 1. f′(x) = 5x⁴ + 3, so f′(1) = 8 and g′(5) = 1/8.
f is decreasing, f(4) = 10 and f′(4) = −2. Find (f⁻¹)′(10).
1/f′(4) = −1/2. A decreasing function has a decreasing inverse.
What happens to the inverse where f′(a) = 0?
The inverse has a vertical tangent at (b, a) and is not differentiable there. Example: f(x) = x³ has f′(0) = 0, so ∛x has no derivative at 0.
What is the most common mistake with inverse derivatives?
Using 1/f′(b) instead of 1/f′(a). Find a with f(a) = b first. Also, f⁻¹ does not mean 1/f.

3.4 Differentiating Inverse Trigonometric Functions

What are the derivatives of arcsin x, arccos x and arctan x?
arcsin x → 1/√(1 − x²); arccos x → −1/√(1 − x²) (both for x between −1 and 1); arctan x → 1/(1 + x²) for all x.
How is d/dx arcsin x = 1/√(1 − x²) derived?
y = arcsin x means sin y = x. Implicitly, cos y·y′ = 1, so y′ = 1/cos y = 1/√(1 − x²). The root is positive because cos y ≥ 0 for y in [−π/2, π/2].
Differentiate arcsin(3x).
3/√(1 − 9x²). Use u′/√(1 − u²) with u = 3x, and square all of u.
Find the derivative of arctan(x³) at x = 1.
3x²/(1 + x⁶) at x = 1 gives 3/2.
Is arcsin x differentiable at x = 1?
No. The denominator √(1 − x²) is 0 there, and the graph has a vertical tangent. arcsin and arccos are not differentiable at x = ±1; arctan is differentiable everywhere.
Which exact inverse trig values should you know?
arcsin(1/2) = π/6, arccos(1/2) = π/3, arctan 1 = π/4, arctan 0 = 0. Use radians.

3.5 Selecting Procedures for Calculating Derivatives

How do you choose which derivative rule to use first?
Rewrite if it helps (roots and reciprocals to powers, log properties). Then find the last operation you would do when evaluating: it picks the first rule. Repeat on each piece.
Differentiate 4/√x.
Rewrite as 4x^(−1/2) first: the derivative is −2x^(−3/2). No quotient rule needed.
Find the derivative of (x² + 1)⁵ at x = 1.
Chain rule: 5(x² + 1)⁴·2x. At x = 1: 5 × 16 × 2 = 160.
Which rule comes first for sin(x²eˣ)?
The chain rule, because sin is applied last. The inside x²eˣ then needs the product rule.
Differentiate eˣ sin x.
Product rule: eˣ sin x + eˣ cos x = eˣ(sin x + cos x).
How do you differentiate with table values?
Write the rule with letters first, then substitute. For f(g(x)) at x = 1, use f′(g(1))·g′(1), not f′(1)·g′(1).

3.6 Calculating Higher-Order Derivatives

What does d²y/dx² mean?
The second derivative: d/dx (dy/dx), also written f″(x) or y″. It is not (dy/dx)².
What are the derivative patterns for e^(kx) and sin x?
The nth derivative of e^(kx) is kⁿe^(kx). sin x cycles: cos x, −sin x, −cos x, sin x, repeating every 4. For sin(kx), each step also adds a factor k.
Let f(x) = x sin x. Find f″(x).
f′(x) = sin x + x cos x. f″(x) = cos x + cos x − x sin x = 2 cos x − x sin x.
What is the 10th derivative of x⁹ + 4x?
0. A degree-9 polynomial has a constant 9th derivative, so every later derivative is 0.
Does f″(a) exist whenever f′(a) exists?
No. f″(a) needs f′ to be differentiable at a. Example: x^(4/3) has f′(0) = 0, but f′(x) = (4/3)x^(1/3) has a vertical tangent at 0, so f″(0) does not exist.
How do you find a second derivative implicitly?
Differentiate dy/dx again, remembering y depends on x (so y terms get y′). Substitute the expression for dy/dx, then simplify using the curve's equation.
Find d²/dx² [e^(5x)] and y‴ for y = x⁶.
25e^(5x); y′ = 6x⁵, y″ = 30x⁴, y‴ = 120x³.

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