IB Maths AA SL
Topic 5 — Calculus
Paper 1 & 2
~6 min read
Second Order Derivatives
A second derivative is what you get when you differentiate twice. That’s literally it. The first derivative tells you the gradient; the second tells you how the gradient itself is changing. It’s the key tool for spotting maximums, minimums, and concavity in upcoming notes.
📘 What you need to know
- Just differentiate twice. No new rules — same chain, product, quotient, etc.
- Notation: if y = f(x), then first derivative = dydx = f′(x), second derivative = d2ydx2 = f″(x).
- It’s used to test stationary points (max or min), find concavity, and locate points of inflection.
- Watch the negative/fractional powers — sign errors are the #1 mistake.
The notation
function
y = f(x)
the original
1st derivative
dy/dx = f′(x)
gradient
2nd derivative
d²y/dx² = f″(x)
how gradient changes
The “²” sits in different places: d² on top, x² on bottom. It’s not d × d, it just means “differentiate twice with respect to x twice”. The notation looks weird but it’s just a label.
What does it actually tell you?
| If you know… | You can find… |
|---|
| f″(x) > 0 | Curve is concave up (smiley shape) → stationary point is a local minimum |
| f″(x) < 0 | Curve is concave down (frowny shape) → stationary point is a local maximum |
| f″(x) = 0 | Possible point of inflection (need to check concavity changes) |
🧠“Position → Velocity → Acceleration”
If f(x) is position, f′(x) is velocity (rate of change of position), and f″(x) is acceleration (rate of change of velocity). Same idea, different layer.
How to find it
1. Differentiate f(x) → get f′(x) 2. Differentiate again → get f″(x)
📍Rewrite first if needed
For functions with roots or fractions, rewrite as powers of x before differentiating — same trick as before. Sign errors with negative/fractional powers are everywhere on this topic.
Worked examples
Find d²y/dx² for y = 2x⁴ − 3x³ + 5x − 7.
first derivative
dy/dx = 8x³ − 9x² + 5second derivative
d²y/dx² = 24x² − 18x
d²y/dx² = 24x² − 18x
just differentiate twice — easy when there are no fractions or roots!
WE 2Roots and fractions — rewrite first
Given f(x) = 4 − √x + 3/√x, (a) find f′(x) and f″(x). (b) Evaluate f″(3) in the form a√b.
part (a) — rewrite first
f(x) = 4 − x^(1/2) + 3x^(−1/2)
f′(x): −(1/2)x^(−1/2) − (3/2)x^(−3/2)
f″(x) (differentiate again):
f″(x) = (1/4)x^(−3/2) + (9/4)x^(−5/2)part (b) — sub x = 3
Rewrite back: f″(x) = 1/(4x√x) + 9/(4x²√x)
f″(3) = 1/(12√3) + 9/(36√3) = 12/(36√3) = 1/(3√3)
Rationalise: × √3/√3 = √3/9
f″(3) = (1/9)√3
two negatives in the second derivative → both terms become positive!
Find d²y/dx² for y = sin(3x).
first derivative
dy/dx = 3 cos(3x)second derivative
Differentiate 3 cos(3x):
d²y/dx² = 3 × (−3 sin(3x)) = −9 sin(3x)
d²y/dx² = −9 sin(3x)
notice: f″(x) = −9 f(x). Trig functions often “loop back” like this!
Find f″(x) for f(x) = e^(2x) + ln x.
first derivative
f′(x) = 2 e^(2x) + 1/x = 2 e^(2x) + x^(−1)second derivative
2 e^(2x) → 4 e^(2x)
x^(−1) → −x^(−2) = −1/x²
f″(x) = 4 e^(2x) − 1/x²
always rewrite 1/x as x^(−1) before differentiating again!
WE 5With chain rule — find f″(0)
For f(x) = (2x + 1)⁴, find f″(0).
first derivative — chain rule
f′(x) = 4(2x + 1)³ × 2 = 8(2x + 1)³second derivative — chain rule again
f″(x) = 8 × 3(2x + 1)² × 2 = 48(2x + 1)²sub x = 0
f″(0) = 48(0 + 1)² = 48 × 1
f″(0) = 48
two chain rules — the inside derivative (2) appears twice → multiplies in!
💡 Top tips
- Rewrite roots and fractions as powers of x before each differentiation.
- Track signs carefully with negative powers — n − 1 with n = −1/2 gives −3/2, not −1/2.
- Use chain/product/quotient rule if needed for either step. Same rules, just twice.
- Convert back to roots or fractions if the question asks for an exact form.
- Notation matters — d²y/dx² and f″(x) mean the same thing. Match the question’s style.
⚠ Common mistakes
- Sign errors with negative powers in the second derivative. (−1/2)(−3/2) = +3/4, not −3/4.
- Forgetting chain rule on the second go. (2x+1)³ → 3(2x+1)² × 2, not just 3(2x+1)².
- Stopping after the first derivative when the question asks for the second.
- Mixing up f′(x) and f″(x) when evaluating at a point. Read the question carefully.
- Wrong superscript placement in the notation — d²y/dx², not dy²/dx².
Now you can differentiate once or twice. Next: stationary points — using f′(x) and f″(x) together to find max and min points on a curve.
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