Something wrapped inside something else — differentiate the outside leaving the inside alone, then multiply by the inside's own derivative.
Differentiate
f(x) = (2x + 1)⁵
Outside: the power rule gives 5(2x + 1)⁴ — then times the inside's derivative, 2: 10(2x + 1)⁴. A forgets the inside factor, the single most common chain-rule error in existence. D differentiates the inside where it should have been left alone.
(2x + 1)⁵ is the function "raise to the fifth" wrapped around the function "2x + 1". Whenever you would say "the quantity, raised / rooted / whatever", a chain rule is coming. Name the inside u and the outside shape, and the rest is bookkeeping.
outside: u⁵ → 5u⁴
power rule, inside untouched
inside: 2x + 1 → 2
its own little derivative
f′(x) = 5(2x + 1)⁴ · 2 = 10(2x + 1)⁴
multiply the layers
(x² + 1)²: chain → 2(x² + 1) · 2x = 4x³ + 4x
the rule's answer
expand: x⁴ + 2x² + 1 → 4x³ + 4x ✓
brute force agrees exactly
Without the inside factor 2x, the chain answer would be 2(x² + 1) = 2x² + 2 — visibly missing the x³ that the expansion demands. The inside derivative is not decoration.
| Composite | Derivative |
|---|---|
| (u)ⁿ | n(u)ⁿ⁻¹ · u′ |
| √u | u′2√u |
| (3x² − 5)⁴ | 4(3x² − 5)³ · 6x = 24x(3x² − 5)³ |
| 1x³ + 2 = (x³ + 2)−1 | −1(x³ + 2)−2 · 3x² = −3x²(x³ + 2)² |
Where is the slope 0? Every factor of f′ can make it zero. For f(x) = (x² − 2)³, f′(x) = 3(x² − 2)² · 2x = 6x(x² − 2)². So 6x = 0 gives x = 0, and x² − 2 = 0 gives x = √2 or x = −√2: three answers.
The inside derivative is negative. Find f′(2) for f(x) = √9 − 4x.
f′(x) = (−4) ÷ (2√9 − 4x)
u′2√u with u′ = −4
= −2√9 − 4x
simplify the 4 and 2
f′(2) = −2√1 = −2
9 − 8 = 1
What is different: the inside falls as x grows, so the slope is negative. Dropping the minus of −4x gives +2.
A cube root. Find f′(2) for f(x) = 3√x² + 23.
f(x) = (x² + 23)1/3
rewrite as a power
f′(x) = ⅓(x² + 23)−2/3 · 2x
⅓ − 1 = −⅔, then × inside′
x = 2: inside = 27, 27−2/3 = 19
(3√27)² = 9
f′(2) = ⅓ × 19 × 4 = 427
multiply the three pieces
What is different: the outside power is a fraction, so the leftover power is negative and the inside ends up below the bar.
A chain inside a product. Find f′(1) for f(x) = x(2x − 1)³.
u = x, u′ = 1; v = (2x − 1)³, v′ = 3(2x − 1)² · 2
v′ needs the chain rule
f′(x) = (2x − 1)³ + 6x(2x − 1)²
u′v + uv′
f′(1) = 1 + 6 = 7
2(1) − 1 = 1
What is different: two rules at once. Product rule on the outside, chain rule for one factor's derivative.
Treating x as a constant and differentiating only the bracket gives 6 at x = 1. The x is a factor that changes too, so it takes its turn in the product rule.
Let h(x) = g(x³), where all we know is g′(8) = 3. Find h′(2).
h′(x) = g′(x³) · 3x²
outside′ with inside kept, × inside′
x = 2: inside = 2³ = 8
that is why we were given g′ at 8
h′(2) = g′(8) · 3(2²) = 3 · 12 = 36
multiply the layers
The given number (here 8) is the inside, evaluated at x = 2. Answering 3 forgets the inside derivative; using g′(2), or 3x² without the square (3 · 2 = 6), are the other classic slips.
Expect a power or root of a polynomial evaluated at a point, a 1(…)ⁿ to rewrite as a negative power, or g(inside) with only numbers given. Every wrong choice is built by dropping the inside derivative — so before choosing, check that your answer contains the inside′ factor.
Differentiate
f(x) = √x² + 9
Outside √u gives 12√u, inside gives 2x, and the 2s cancel: x√x² + 9. B stops after the outside — no inside factor — and C keeps only the inside. Check at x = 4: f′ = 45 = 0.8. Numerically, f(4) = 5 and f(4.01) = √25.0801 ≈ 5.008, a rise of 0.008 over a run of 0.01, so the slope is ≈ 0.8 ✓.
