§ 01What Is a Higher-Order Derivative?
The derivative of a function is itself a function, and a function can be differentiated again. Repeating this process produces second, third, and in general nth derivatives.
When you differentiate f(x), you get f′(x), which tells you the rate of change of f. When you differentiate f′(x), you get f″(x), which tells you the rate of change of the rate of change, in other words, how quickly the slope itself is changing. This is the second derivative, and it governs crucial geometric properties like concavity and inflection points.
You can keep going: f‴(x) is the third derivative, and so on. In mechanics, if f(t) is position, each successive derivative has a vivid physical name: velocity, acceleration, jerk, snap, crackle, pop (yes, those are the real names).
Each arrow means "differentiate with respect to x." Any step can be taken as many times as needed.
§ 02All the Notation Systems
Higher-order derivatives come with three competing notation families. You need to recognise all of them, because different textbooks and fields favour different styles.
2nd: f″(x)
3rd: f‴(x)
4th+: f(4)(x), f(5)(x), …
nth: f(n)(x)
2nd: d²y/dx²
3rd: d³y/dx³
nth: dny/dxn
1st derivative: ẏ (one dot)
2nd derivative: ÿ (two dots)
Rarely used beyond 2nd order
D²f = d²f/dx²
Dnf = dnf/dxn
Useful in differential equations
Quick Translation Table
| Order | Lagrange | Leibniz | Newton (time) | Operator |
|---|---|---|---|---|
| 1st | f′(x) | dy/dx | ẏ | Df |
| 2nd | f″(x) | d²y/dx² | ÿ | D²f |
| 3rd | f‴(x) | d³y/dx³ | ẏ | D³f |
| nth | f(n)(x) | dny/dxn | rarely used | Dnf |
§ 03Physical Meaning — Position to Jerk
The most vivid illustration of higher-order derivatives comes from classical mechanics. If s(t) is the position of a particle at time t, each successive derivative has a concrete physical name and meaning.
Velocity is how fast you are moving. Acceleration is how fast your velocity is changing, what you feel pressed into a seat during take-off. Jerk (also called jolt) is how fast acceleration is changing: the lurch you feel when a lift starts or stops, or the snap of a whiplash. Engineers who design smooth roller coasters and elevator systems work hard to limit jerk.
§ 04Concavity and the Second Derivative
The second derivative gives direct information about the shape (concavity) of a curve, and is the key tool for classifying stationary points.
A curve is concave up where it curves like a bowl facing upward: the tangent lines lie below the curve. It is concave down where it curves like a hill: the tangent lines lie above the curve. An inflection point is where concavity changes sign.
When the test is inconclusive, fall back to the first derivative sign chart.
§ 05The nth Derivative — Patterns for Key Functions
For certain fundamental functions, differentiating repeatedly reveals a beautiful cyclic or algebraic pattern, allowing us to write down the nth derivative directly without performing n steps.
Polynomials — xn Terminates
Each differentiation reduces the power by one and multiplies by the old power. After n steps the result is a constant; one step further gives zero and all subsequent derivatives are zero.
Example: f(x) = x⁵. Then f‴(x) = 60x², f(4)(x) = 120x, f(5)(x) = 120, f(6)(x) = 0.
The Exponential — ex Never Changes
eˣ is its own derivative at every order. No matter how many times you differentiate, you always get eˣ back.
Sine and Cosine — A 4-Cycle
Differentiating sin x and cos x produces a repeating cycle of length 4. After four differentiations you return to the original function.
| n mod 4 | dn/dxn (sin x) | dn/dxn (cos x) |
|---|---|---|
| 0 (n=0,4,8,…) | sin x | cos x |
| 1 (n=1,5,9,…) | cos x | −sin x |
| 2 (n=2,6,10,…) | −sin x | −cos x |
| 3 (n=3,7,11,…) | −cos x | sin x |
Similarly: dn/dxn(cos x) = cos(x + nπ/2). The derivative just phase-shifts the trig function by π/2 each time.
General Power Exponential — eax
Each differentiation multiplies by the constant a. After n steps the factor is aⁿ.
Summary Reference Table
| Function f(x) | nth Derivative f(n)(x) | Key property |
|---|---|---|
| xm | m!/(m−n)! · xm−n (n ≤ m) | Terminates at n=m |
| ex | ex | Self-replicating |
| eax | an eax | Grows by factor a each step |
| sin x | sin(x + nπ/2) | 4-cycle |
| cos x | cos(x + nπ/2) | 4-cycle |
| ln x | (−1)n−1(n−1)!/xn | For n ≥ 1, x > 0 |
| x−1 = 1/x | (−1)n · n!/xn+1 | Alternating sign |
§ 0610 Fully Worked Examples
Examples cover polynomial, trigonometric, exponential, and combined functions, progressing from straightforward repeated differentiation through concavity analysis to general nth derivatives.
Differentiate twice, term by term.
- f′(x) = 5x⁴ − 9x² + 7 (power rule on each term)
- f″(x) = 20x³ − 18x (differentiate f′)
Apply the chain rule on each differentiation. The factor of 2 accumulates.
- dy/dx = 2 cos(2x)
- d²y/dx² = 2 · (−2 sin(2x)) = −4 sin(2x)
- d³y/dx³ = −4 · 2 cos(2x) = −8 cos(2x)
Pattern: each differentiation multiplies by 2. After n steps: 2ⁿ sin(2x + nπ/2).
First derivative requires the product rule; then differentiate again.
- f′(x) = eˣ + x eˣ = (1 + x)eˣ (product rule)
- f″(x) = d/dx[(1+x)eˣ]. Apply the product rule again: (1+x)′·eˣ + (1+x)·eˣ = eˣ + (1+x)eˣ.
- Simplify: eˣ[1 + 1 + x] = (2 + x)eˣ.
Find where f′ = 0, then use the second derivative test.
- f′(x) = 3x² − 3 = 3(x²−1) = 0 → x = ±1.
- f″(x) = 6x.
- At x = 1: f″(1) = 6 > 0 → local minimum. f(1) = 1 − 3 = −2.
- At x = −1: f″(−1) = −6 < 0 → local maximum. f(−1) = −1 + 3 = 2.
Inflection points occur where f″ = 0 and changes sign.
- f′(x) = 4x³ − 12x.
- f″(x) = 12x² − 12 = 12(x²−1) = 0 → x = ±1.
- For x < −1: f″ = 12(4−1) > 0 (concave up). For −1 < x < 1: f″ = 12(0−1) < 0 (concave down). For x > 1: concave up. Sign changes at both x = ±1. ✓
- f(1) = 1 − 6 = −5. f(−1) = 1 − 6 = −5.
Use the 4-cycle pattern: cos x → −sin x → −cos x → sin x → cos x.
- f′(x) = −sin x.
- f″(x) = −cos x.
- f‴(x) = sin x.
- f(4)(x) = cos x. The cycle has completed: we are back to the start.
This confirms: dn/dxn(cos x) = cos(x + nπ/2). At n=4: cos(x + 2π) = cos x. ✓
The chain rule contributes a factor of 3 each time we differentiate.
- f′(x) = 3e3x.
- f″(x) = 3² e3x = 9e3x.
- f‴(x) = 3³ e3x = 27e3x.
- Pattern: each step multiplies by 3. After n steps the coefficient is 3ⁿ.
Use the 4-cycle to avoid doing 50 differentiations.
- 50 ÷ 4 = 12 remainder 2. So n mod 4 = 2.
- From the table: when n mod 4 = 2, dn/dxn(sin x) = −sin x.
- Alternatively: sin(x + 50π/2) = sin(x + 25π) = sin(x + π) = −sin x. ✓
- f′(x) = 1/x = x−1.
- f″(x) = d/dx(x−1) = −x−2 = −1/x².
- Since f″(x) = −1/x² < 0 for all x > 0, ln x is concave down everywhere on its domain.
- Velocity: v(t) = s′(t) = 4t³ − 12t². At t=2: v(2) = 32 − 48 = −16.
- Acceleration: a(t) = s″(t) = 12t² − 24t. At t=2: a(2) = 48 − 48 = 0.
- Jerk: j(t) = s‴(t) = 24t − 24. At t=2: j(2) = 48 − 24 = 24.
The zero acceleration at t = 2 means the particle is at a velocity turning point, its speed has momentarily stopped changing, even though jerk ≠ 0 means that will change immediately after.
§ 07Common Mistakes
§ 0810-Question Practice Quiz
Test your command of higher-order derivatives. A mix of computation, classification, and pattern recognition.
If f(x) = x⁴, what is f″(x)? Enter the coefficient: f″(x) = __ x²
- f′(x) = 4x³
- f″(x) = 12x²
What does f″(c) < 0 tell you about the curve at x = c?
- f″(c) < 0 means the slope is decreasing through c: the curve bends downward (∩), i.e. concave down.
- A local minimum requires f′(c)=0 AND f″(c)>0, two conditions together.
Find d²y/dx² for y = e5x. It equals __ · e5x.
- dy/dx = 5e5x
- d²y/dx² = 25e5x
What is f(3)(x) for f(x) = cos x?
- f′ = −sin x, f″ = −cos x, f‴ = sin x.
- Or: cos(x + 3π/2) = cos x cos(3π/2) − sin x sin(3π/2) = 0 + sin x = sin x. ✓
For f(x) = x⁶, what is f(6)(x)? (It's a constant: enter the value.)
- dn/dxn(xn) = n! for the nth derivative.
- 6! = 6×5×4×3×2×1 = 720.
A particle's position is s(t) = t³ − 6t. What is its acceleration at t = 2?
- s′(t) = 3t² − 6 (velocity)
- s″(t) = 6t (acceleration)
- s″(2) = 12
Find f(100)(x) for f(x) = sin x. Enter the number k where f(100)(x) = k·sin x.
- 100 ÷ 4 = 25 remainder 0. So n mod 4 = 0.
- When n mod 4 = 0, dn/dxn(sin x) = sin x.
- So f(100)(x) = sin x, meaning k = 1.
The second derivative test is inconclusive when:
- The second derivative test requires: f′(c) = 0 (critical point) AND the sign of f″(c) to classify it.
- If both are zero, the test gives no information, use the first derivative sign chart instead.
Find f‴(x) for f(x) = x⁵. Enter the coefficient of x²: f‴(x) = __ x²
- f′(x) = 5x⁴
- f″(x) = 20x³
- f‴(x) = 60x²
Which of these is the correct Leibniz notation for the second derivative of y with respect to x?
- d²y/dx² is the Leibniz notation for "apply d/dx twice to y."
- (dy/dx)² is the square of the first derivative, a completely different quantity.
Quiz complete!