Calculus I › Differentiation rules › full formula sheet
Higher-order derivatives
Differentiate the derivative: the second derivative measures how the slope itself changes — concavity, acceleration, and beyond.
Notation: f″ = second derivative, f″′ = third; Leibniz writes d²y/dx², d³y/dx³. f(n) = nth derivative.
Before this lesson: Power rule · Derivative of sin x
Where it comes from
The first derivative is velocity: how position changes. But velocity changes too — a car accelerates. The second derivative is the derivative of the derivative: the rate of change of the rate of change. And nothing stops there: differentiate again for the third, again for the fourth.
Geometrically: f′ tells you whether the graph rises or falls; f″ tells you whether the slope itself is rising or falling — that’s concavity (cup up vs. cup down). A line has f″ = 0 everywhere: its slope never changes.
The wrong guess to kill: f″(x) = (f′(x))²?? Take f(x) = x³: (f′)² = (3x²)² = 9x⁴; but f″ = d/dx [3x²] = 6x. At x = 1: 9 ≠ 6. Dead on arrival — the second derivative differentiates again; it doesn’t square.
Before reading on: f(x) = x³. Is f″(x) = (f′(x))² = 9x⁴, or does “differentiate again” give something else? Test both at x = 1.
Derivation
There’s nothing to prove — higher derivatives are defined by iterating the derivative. What needs explaining is the notation, especially Leibniz’s:
Before reading on: d/dx applied twice to y — if each application adds one d on top and one dx below, what notation would you invent for the result?
Prime vs. Leibniz: f″ is compact (good for formulas); d²y/dx² names the variable (good for physics and PDEs). f(n) takes over past the third prime — f″′′ gets silly.
How to use it
The procedure couldn’t be simpler: differentiate, then differentiate the answer.
- Find f′ with the usual rules.
- Differentiate f′ to get f″ — same rules, new function.
- Repeat for f″′, f(4), …
- Interpret: f″ > 0 ⇒ concave up (cup ∪, slope increasing); f″ < 0 ⇒ concave down (cap ∩, slope decreasing); f″ = 0 ⇒ possible inflection.
Judgment calls
Physics: s(t) → v(t) = s′ → a(t) = s″. Acceleration is the second derivative of position — that’s why f″ matters beyond curve sketching. Don’t square: f″ means “differentiate twice,” never (f′)².
Worked examples
Four problems, easiest first. Climb the ladder one rung at a time.
Example 1 — f(x) = x⁴: find f″ and f″′
- First: f′(x) = 4x³.
- Second: f″(x) = 12x².
- Third: f″′(x) = 24x. (Why keep going? Each rung is one more power-rule application.)
- Fourth: f(4)(x) = 24 — and f(5) = 0. Polynomials always bottom out.
Your turn: f(x) = x⁵: find f″ and f″′
Answer: f″(x) = 20x³, f″′(x) = 60x²
First: f′(x) = 5x⁴. Second: f″(x) = 20x³. Third: f″′(x) = 60x². One more round: f(4)(x) = 120x, then 120, then 0 — polynomials always bottom out.
Example 2 — f(x) = sin x: the 4-cycle
- f′ = cos x.
- f″ = −sin x.
- f″′ = −cos x. f(4) = sin x — back where we started.
- The cycle: sin → cos → −sin → −cos → sin … (Why? Each differentiation shifts the wave; four shifts = full period.)
Your turn: f(x) = cos x: the 4-cycle
Answer: f″(x) = −cos x, f″′(x) = sin x
f′ = −sin x. f″ = −cos x. f″′ = sin x. f(4) = cos x — back where we started: cos → −sin → −cos → sin → cos …
Example 3 — f(x) = e2x: find f″
- First: f′(x) = 2e2x (chain: ×2).
- Second: f″(x) = 2·2e2x = 4e2x. (Why 4? Each round multiplies by the chain factor 2 again.)
- Pattern: f(n)(x) = 2n·e2x — exponentials never bottom out.
Your turn: f(x) = e3x: find f″
Answer: 9e3x
First: f′(x) = 3e3x (chain: ×3). Second: f″(x) = 3·3e3x = 9e3x. The ×3 applies every round — f(n)(x) = 3n·e3x.
Example 4 — physics: s(t) = t³ − 6t² (position in meters)
- Velocity: v(t) = s′(t) = 3t² − 12t.
- Acceleration: a(t) = s″(t) = 6t − 12.
- At t = 2: a(2) = 0 — instantaneously not accelerating (velocity momentarily extremal). At t = 3: a(3) = 6 m/s².
- Meaning: a > 0 for t > 2 — velocity increasing after t = 2. The second derivative is the velocity’s slope ✓
Your turn: Physics: s(t) = t⁴ − 8t³ (position in meters)
Answer: a(4) = 0 m/s²
Velocity: v(t) = s′(t) = 4t³ − 24t². Acceleration: a(t) = s″(t) = 12t² − 48t. At t = 4: a(4) = 12·16 − 48·4 = 192 − 192 = 0 — velocity momentarily extremal.
Memorization tips
- Count the primes: two primes, two differentiations. The #1 error is stopping one rung early.
- Never square: f″ means “differentiate twice,” not (f′)². The x³ test (6x vs 9x⁴) kills the confusion.
- The sin 4-cycle: sin → cos → −sin → −cos → sin. Memorize the wheel — any-order trig derivatives become instant.
- Leibniz decodes: d²y/dx² = d/dx(dy/dx). Read it as “differentiate dy/dx” and the notation explains itself.
- Polynomials bottom out: x⁴ → 0 by the 5th derivative. Exponentials never do: (e2x)(n) = 2ne2x.
- Physics ladder: position → velocity → acceleration. f″ > 0 means speeding up in the velocity sense — the slope is increasing.
Final challenge
Five mixed questions — the sin cycle, acceleration, and notation. Score 5/5 and it’s yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the second derivative?
The derivative of the derivative: f′′(x) = d/dx[f′(x)], written f′′ or d²y/dx². It measures how the slope itself changes — concavity geometrically, acceleration physically.
What does d²y/dx² actually mean?
Apply d/dx twice: d/dx[dy/dx]. The d’s “multiply” to d² and dx·dx becomes dx². Note d²y is not (dy)² — it’s notation for “the second differential.”
Is f′′(x) the same as (f′(x))²?
No! For f(x) = x³: f′′ = 6x but (f′)² = 9x⁴. The second derivative differentiates again; squaring is a completely different operation.
What are the higher derivatives of sin x?
They cycle every 4: sin → cos → −sin → −cos → sin… So f′ = cos x, f′′ = −sin x, f′′′ = −cos x, f⁽⁾ = sin x.
What is acceleration in calculus terms?
The second derivative of position: if s(t) is position, v(t) = s′(t) is velocity and a(t) = s′′(t) is acceleration. For s(t) = t³−6t², a(t) = 6t−12.
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