IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
~6 min read
Higher Order Derivatives
Differentiate once you get f′(x); differentiate again you get f″(x) — the second derivative. Keep going for f‴, f⁽⁴⁾, f⁽⁵⁾, … The second derivative tells you about concavity (next note) and helps classify stationary points. Higher derivatives often follow a pattern, especially for sin/cos/exp.
📘 What you need to know
- Second derivative: f″(x) = d²y/dx² — differentiate the first derivative once more.
- Notation: dash for the first three (f′, f″, f‴), then bracketed numbers (f⁽⁴⁾, f⁽⁵⁾, …); equivalent forms d²y/dx², d³y/dx³, etc.
- Position of superscript 2’s: d²y/dx² — differentiating twice (d²) with respect to x twice (x²).
- Polynomials eventually have higher derivatives equal to zero (each derivative drops the power by 1).
- ekx follows a clean pattern: f⁽ⁿ⁾(x) = kn ekx.
- sin/cos cycle every 4 derivatives — sin → cos → −sin → −cos → sin. Useful for “find f⁽ⁿ⁾” pattern problems.
- Uses: classify stationary points (next note), determine concavity, sketch derivatives, prove identities.
- GDC trick: most calculators only return f′ and f″ directly, but you can differentiate twice algebraically and then use the result as the “original” function on the GDC for the third or fourth derivative.
Notation
| Order | Function notation | Leibniz notation |
|---|
| 1st | f′(x) | dy/dx |
| 2nd | f″(x) | d²y/dx² |
| 3rd | f‴(x) or f⁽³⁾(x) | d³y/dx³ |
| 4th, 5th, … | f⁽⁴⁾(x), f⁽⁵⁾(x), … | d⁴y/dx⁴, d⁵y/dx⁵, … |
| n-th | f⁽ⁿ⁾(x) | dⁿy/dxⁿ |
Beyond third order: three or more dashes get unreadable, so use parentheses with the order number — f⁽⁴⁾, f⁽⁵⁾, … The bracket is to distinguish from a power: f⁴(x) without brackets often means [f(x)]⁴ in some texts.
Standard patterns for higher derivatives
Exponential ekx
f⁽ⁿ⁾(x) = kn ekx
multiply by k each time — the ekx is preserved
Sine cycle (period 4)
sin → cos → −sin → −cos → sin
every 4 derivatives sin returns to itself
Cosine cycle (period 4)
cos → −sin → −cos → sin → cos
same 4-cycle, started one step ahead
Polynomial of degree n
f⁽ⁿ⁺¹⁾(x) = 0
degree drops by 1 each time, eventually reaching zero
🧭 Recipe — find a higher-order derivative
- Rewrite roots, fractions, and reciprocals as powers of x first (e.g. 1/x² = x⁻², √x = x^½).
- Differentiate once using the appropriate rule (power, chain, product, quotient).
- Repeat for each additional order required — keep working carefully through negative/fractional powers.
- For pattern problems: compute the first three or four derivatives and spot the pattern (coefficient, sign, function).
- Evaluate at any required x-value and rationalise the denominator if asked for an exact form.
Worked examples
WE 1Polynomial — f′ and f″
Given f(x) = 2x⁴ − 5x³ + 7x − 3, find f′(x) and f″(x).
Differentiate once (power rule term-by-term)
f′(x) = 8x³ − 15x² + 7
Differentiate again
f″(x) = 24x² − 30x
f′(x) = 8x³ − 15x² + 7; f″(x) = 24x² − 30x
the constant −3 vanishes in f′; the +7 vanishes in f″. Each differentiation drops one term
WE 2Roots and reciprocals — evaluate f″(4)
f(x) = 6√x − 4/x². (a) Find f′(x) and f″(x). (b) Evaluate f″(4).
(a) Rewrite as powers of x
f(x) = 6 x^(1/2) − 4 x^(−2)
Differentiate once
f′(x) = 3 x^(−1/2) + 8 x^(−3)
= 3/√x + 8/x³
Differentiate again
f″(x) = −(3/2) x^(−3/2) − 24 x^(−4)
= −3/(2x√x) − 24/x⁴
(b) Substitute x = 4
f″(4) = −3/(2·4·2) − 24/256
= −3/16 − 24/256 = −48/256 − 24/256
= −72/256 = −9/32
(a) f″(x) = −3/(2x√x) − 24/x⁴; (b) f″(4) = −9/32
careful with −1/2 + (−1) = −3/2: when you reduce the power on x^(−1/2), you go MORE negative
WE 3Trig combination — second derivative
Find f″(x) for f(x) = cos(3x) + sin(2x).
Differentiate once
f′(x) = −3 sin(3x) + 2 cos(2x)
Differentiate again
f″(x) = −3 · 3 cos(3x) + 2 · (−2) sin(2x)
= −9 cos(3x) − 4 sin(2x)
f″(x) = −9 cos(3x) − 4 sin(2x)
notice f″ = −9 cos(3x) − 4 sin(2x) and f = cos(3x) + sin(2x) — they’re not simply related, because the two terms have different periods
WE 4Exponential — pattern for the n-th derivative
For f(x) = e2x: (a) Compute f′, f″, f‴ and find a formula for f⁽ⁿ⁾(x). (b) Hence write down f⁽⁵⁾(x).
(a) Compute first three derivatives
f′(x) = 2 e^(2x)
f″(x) = 4 e^(2x)
f‴(x) = 8 e^(2x)
Spot the pattern: each derivative multiplies by 2
f^(n)(x) = 2^n · e^(2x)
(b) Apply for n = 5
f^(5)(x) = 2⁵ · e^(2x) = 32 e^(2x)
(a) f⁽ⁿ⁾(x) = 2ⁿ e^(2x); (b) f⁽⁵⁾(x) = 32 e^(2x)
the ekx never goes away — each derivative just multiplies by another k
WE 5Second derivative needing product rule
Given y = x² ln(x), find d²y/dx².
First derivative — product rule
u = x², v = ln x; u′ = 2x, v′ = 1/x
dy/dx = x²·(1/x) + ln(x)·2x = x + 2x ln(x)
= x(1 + 2 ln x)
Second derivative — differentiate term-by-term
d/dx[x] = 1
d/dx[2x ln x] = 2 ln x + 2x·(1/x) = 2 ln x + 2 (product rule again)
d²y/dx² = 1 + 2 ln x + 2 = 2 ln x + 3
d²y/dx² = 2 ln(x) + 3
tidy: by x = 1, d²y/dx² = 3, suggesting (1, 0) on the curve is a concave-up point — check next note
WE 6Show-that — sin satisfies a harmonic relation
Show that f(x) = sin(3x) satisfies f″(x) + 9 f(x) = 0.
Differentiate twice
f′(x) = 3 cos(3x) (chain rule)
f″(x) = −9 sin(3x) (chain rule again)
Substitute into the LHS
f″(x) + 9 f(x) = −9 sin(3x) + 9 sin(3x)
= 0 ✓
f″(x) + 9 f(x) = 0 ∎
this is the simple harmonic motion equation — sin(kx) and cos(kx) both satisfy f″ + k² f = 0; useful in Physics IA
💡 Top tips
- Convert before differentiating — rewrite roots, fractions, reciprocals as powers of x; otherwise mistakes pile up at the second derivative.
- Pattern problems — write out f′, f″, f‴ explicitly. The pattern usually shows up by the 3rd derivative (sign cycle, coefficient pattern, function cycle).
- For “show that” identities involving f and f″ — work each side separately and show they’re equal; don’t try to manipulate the equation as a whole.
- GDC for higher than f″ — differentiate twice algebraically to get a “new function” g(x); then use GDC’s d/dx on g to evaluate f‴ or f⁽⁴⁾.
- Sin/cos cycle: every 4 derivatives, the function returns to itself (with the chain factor compounded).
⚠ Common mistakes
- Power errors — when reducing x^(−1/2) by 1, you get x^(−3/2), NOT x^(1/2). Subtract 1 from the exponent.
- Sign mistakes on chain trig — d/dx(cos(3x)) = −3 sin(3x); going to the second derivative, d/dx(−3 sin(3x)) = −9 cos(3x) (sign survives, factor of 3 multiplies in).
- Notation confusion — d²y/dx² has the 2 in TWO places (d² on top, dx² on bottom). Don’t write dy²/dx².
- Forgetting product rule when differentiating f′ — if f′(x) is itself a product (like 2x ln x), use product rule again on it.
- Stopping too early on patterns — sometimes the pattern only emerges at the 4th derivative (especially for sin/cos cycles).
Up next: Stationary Points. Now that you can compute f′ and f″, you’ll classify stationary points (where f′ = 0) as local minima, local maxima, or points of inflection. The second derivative test: f″(x) > 0 → minimum; f″(x) < 0 → maximum; f″(x) = 0 → inconclusive (use the first-derivative sign test instead).
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