AP Calc AB The Chain RuleAP Calc AB The Chain Rule

APCALCULUS · UNIT3

The Chain Rule

Topic 3.4 Study Guide

In one sentence: to differentiate a function inside another function, take the derivative of the outside (leaving the inside alone), then multiply by the derivative of the inside.

The rule

d/dx [ f(g(x)) ] = f ′(g(x)) · g′(x)

Why it works: if y changes 3 times as fast as u, and u changes 2 times as fast as x, then y changes 6 times as fast as x. Rates multiply.

Four steps

1. Spot the outside function (last thing you would do when evaluating) and the inside. 2. Differentiate the outside, copy the inside unchanged.

3. Multiply by the derivative of the inside.

4. Simplify (factor, remove negative exponents).

Patterns to memorize

Function Derivative

[g(x)]ⁿ n[g(x)]ⁿ⁻¹ · g′(x)

√g(x) g′(x) / (2√g(x))

sin g(x) cos g(x) · g′(x)

cos g(x) −sin g(x) · g′(x)

tan g(x) sec² g(x) · g′(x)

sec g(x) sec g(x) tan g(x) · g′(x)

csc g(x) −csc g(x) cot g(x) · g′(x)

Trap. The classic error is forgetting the inside derivative. Also, cos 4x gives −4 sin 4x, so the sign flips and you multiply by 4.

Worked examples

Powers and roots

1. g(x) = 3(4 − 9x)⁴. Outside: 3(·)⁴. Inside: 4 − 9x. g′ = 3·4(4 − 9x)³·(−9) = −108(4 − 9x)³. 2. h(x) = √(5 − x) = (5 − x)^½. h′ = ½(5 − x)^(−½)(−1) = −1 / (2√(5 − x)).

3. y = ∛(6x² + 1). y′ = ⅓(6x² + 1)^(−⅔)(12x) = 4x / (6x² + 1)^(2/3).

4. y = 1/√(3x + 5) = (3x + 5)^(−½). y′ = −½(3x + 5)^(−3/2)(3) = −3 / (2(3x + 5)^(3/2)). Trig inside and trig powers

1. y = cos 4x. y′ = −4 sin 4x.

2. g(x) = 5 tan 3x. g′ = 5 sec²(3x)·3 = 15 sec² 3x.

3. y = 4 sec² x = 4(sec x)². y′ = 4·2 sec x · (sec x tan x) = 8 sec² x tan x. 4. y = tan² 5x = (tan 5x)². y′ = 2 tan 5x · sec² 5x · 5 = 10 tan 5x sec² 5x.

5. y = sin x³. Inside is x³: y′ = 3x² cos x³. Compare with sin³ x whose derivative is 3 sin² x cos x. Read it out loud. tan² 5x means (tan(5x))², so there are two layers (square, then tan) plus the 5x inside, which means three factors in the answer.

Chain rule with the product and quotient rules

1. f(x) = x³√(x + 1). Product rule: f′ = 3x²√(x + 1) + x³ · 1/(2√(x + 1)), which simplifies to x²(7x + 6) / (2√(x + 1)).

2. f(x) = (2x − 3)³ / (x + 2). Quotient rule: f′ = [6(2x − 3)²(x + 2) − (2x − 3)³] / (x + 2)². Factor out (2x − 3)²: (2x − 3)²[6x + 12 − 2x + 3] = (2x − 3)²(4x + 15). So f′ = (2x − 3)²(4x + 15) / (x + 2)².

3. f(x) = ∛(x² + 6x + 8). f′ = (2x + 6) / (3(x² + 6x + 8)^(2/3)) = 2(x + 3) / (3(x² + 6x + 8)^(2/3)).

Practice (no calculator)

1. y = (3x⁴ + 5)³

2. y = √(6x − 2)

3. f(x) = −2/(3 − x²)³

4. f(x) = tan² 5x

5. y = csc 8x

6. y = cos² 2x

7. f(x) = x² sin 3x

8. y = 2 ⁴√(9 − x²)

Answer key

1. y′ = 3(3x⁴ + 5)²(12x³) = 36x³(3x⁴ + 5)²

2. y′ = 3 / √(6x − 2)

3. f′ = −12x / (3 − x²)⁴

4. f′ = 10 tan 5x sec² 5x

5. y′ = −8 csc 8x cot 8x

6. y′ = 2cos 2x · (−sin 2x) · 2 = −4 cos 2x sin 2x

7. f′ = 2x sin 3x + 3x² cos 3x = x(2 sin 3x + 3x cos 3x)

8. y′ = −x / (9 − x²)^(3/4)