IB Maths AI HL
Further Differentiation
Paper 1 & 2
~6 min read
Second Order Derivatives
Differentiate once and you get the gradient. Differentiate the gradient and you get the second derivative — the rate of change of the rate of change. It tells you how the gradient itself is shifting, which is exactly what you need to classify stationary points and read a curve’s concavity in the topics ahead.
📘 What you need to know
- Second derivative = differentiate the function twice.
- Two notations: d²ydx² and f′′(x) — note where the 2’s sit.
- Meaning: the rate of change of the gradient.
- Used to: test stationary points, determine concavity, and help graph f′.
- Method: just differentiate again — often rewriting roots/fractions as powers first.
- Take care with negative and fractional powers on the second pass.
What the notation means
The first derivative is the gradient; the second is the gradient’s gradient. Watch where the superscript 2’s go — they sit differently in the two parts of the fraction.
First and second derivative notation
dydx = f′(x) (first) · d²ydx² = f′′(x) (second)
✗ notation to recognise, not a booklet formula
🧠 “d-two-y over d-x-squared”
You differentiate twice (so d²) with respect to x twice (so x²). The 2 sits up top with the d, but down below with the x — that asymmetry is the thing to memorise.
| Derivative | What it measures |
|---|
| f(x) | the value of the function |
| f′(x) (first) | the gradient — the rate of change of f |
| f′′(x) (second) | the rate of change of the gradient |
Why it’s useful: the second derivative lets you test for local minimum and maximum points, decide the nature of stationary points, determine the concavity of a curve, and help sketch the graph of the derivative — all coming up in the next two topics.
Finding a second derivative
🧭 Recipe — second derivative
- Rewrite roots and fractions as negative/fractional powers of x.
- Differentiate once to get f′(x) — applying chain/product/quotient rules as needed.
- Differentiate again to get f′′(x), working carefully term by term.
- Evaluate or simplify if asked — e.g. tidy surds by rationalising the denominator.
🤔 Why are negative powers so error-prone here?
Each differentiation drops the power by 1, so a term like x−1/2 becomes x−3/2, then x−5/2 — the exponents get more negative and the coefficients pick up extra factors each time. Two passes means twice the chance to slip a sign or mishandle the fraction, so the safest approach is one clean term at a time.
Worked examples
Throughout, f(x) = 4 − √x + 3√x.
WE 1Rewrite f(x) as powers of x
Convert the roots before differentiating.
√x = x^(1/2), 3√x = 3x^(−1/2)
f(x) = 4 − x^(1/2) + 3x^(−1/2)
Differentiate once, term by term.
4 → 0
−x^(1/2) → −½x^(−1/2)
3x^(−1/2) → 3·(−½)x^(−3/2) = −32x^(−3/2)
f′(x) = −½x^(−1/2) − 32x^(−3/2)
Differentiate f′(x) again. Watch the negatives.
−½x^(−1/2) → −½·(−½)x^(−3/2) = ¼x^(−3/2)
−32x^(−3/2) → −32·(−32)x^(−5/2) = 94x^(−5/2)
f″(x) = ¼x^(−3/2) + 94x^(−5/2)
WE 4Evaluate f′′(3) in the form a√b
Write as surds, substitute x = 3, then rationalise.
f″(x) = 14x√x + 94x²√x
f″(3) = 112√3 + 936√3 = 1236√3 = 13√3
rationalise: 13√3 × √3√3 = √39
f″(3) = 19√3
WE 5Find f′′(x) for f(x) = 2x3 − 5x2 + 4x − 1
A polynomial — straightforward double differentiation.
f′(x) = 6x² − 10x + 4
f″(x) = 12x − 10
f″(x) = 12x − 10
💡 Top tips
- Rewrite first — convert roots and fractions to powers before differentiating.
- Go term by term — second derivatives are where careless slips happen.
- Mind the 2’s in the notation: d²ydx² (up with d, down with x).
- Rationalise surds when a question wants the form a√b.
- Apply the right rule each pass — chain/product/quotient may be needed both times.
- Sanity-check the sign — it’ll tell you concavity later, so it matters.
⚠ Common mistakes
- Power-rule slips on the second pass — x−1/2 → x−3/2, not x−1/2.
- Sign errors — two negatives multiply to a positive (e.g. −½·−½ = +¼).
- Misplacing the 2’s — it’s d²ydx², not dy²dx².
- Stopping at f′ — read carefully; the question wants the second derivative.
- Leaving surds un-rationalised when a specific form is requested.
Next up — Stationary Points. With both first and second derivatives in hand, you can now do more than just find stationary points (where f′(x) = 0) — you can classify them. The next topic uses the sign of f′′(x) to decide instantly whether a turning point is a local minimum, a local maximum, or something subtler.
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