IB Maths AI HLDifferentiationPaper 1 & 2~8 min read
Differentiating Powers of x
The chord-by-chord method gave you the idea of a gradient; this is the shortcut. One booklet rule — f(x) = xn → f′(x) = nxn−1 — lets you write down the whole derivative in one line. The only real skill is tidying every term into a power of x first: roots become fractional powers, fractions become negative powers, products and quotients get expanded.
📘 What you need to know
Power rule: f(x) = xn → f′(x) = nxn−1, valid for any n ∈ ℚ. In the booklet.
Constant multiple: f(x) = axn → f′(x) = anxn−1.
Two special cases: ax → a, and a constant a → 0.
Sums & differences: differentiate term by term.
Roots → fractional powers (e.g. √x = x1/2) before differentiating.
Fractions → negative powers (e.g. 4x = 4x−1) before differentiating.
Products & quotients: you can’t differentiate factor-by-factor — expand/simplify first.
The power rule
Differentiation is just the process of finding the derivative (gradient function) from the function. For powers of x there’s a single mechanical rule: bring the power down to the front, then knock one off the power.
Power rule (with constant multiple)f(x) = axn → f′(x) = anxn−1In the formula booklet (n ∈ ℚ) ✓
Multiply down, subtract one
The old power n multiplies the front; the new power is n − 1.
🤔 Why “minus one”?
A derivative is a gradient — a “per unit x” rate. Differentiating drops the degree of every term by exactly one, which is why x3 (a volume-like growth) becomes x2 (an area-like rate), and a straight line ax (degree 1) becomes the constant a (degree 0). A flat constant has no slope at all, so it differentiates to 0.
special case 1ax → aA straight line’s gradient is just its coefficient. y = 6x → dydx = 6.
special case 2a → 0A constant doesn’t change, so its gradient is zero. y = 5 → dydx = 0.
Rewrite first: roots and fractions
The rule only works once every term is an explicit power of x. Two rewrites cover almost everything you’ll meet.
You see
Rewrite as
Then differentiate
2√x
2x1/2
x−1/2
4x
4x−1
−4x−2
4√x
4x−1/2
−2x−3/2
🧭 Recipe — differentiate any power expression
Tidy each term into the form axn: turn roots into fractional powers and fractions into negative powers.
Expand any products or brackets so you have a pure sum/difference of powers.
Apply the rule term by term: front × power, then subtract 1 from each power.
Take care with negatives/fractions: e.g. −12 − 1 = −32. Leave the answer in matching power form.
🧠 “Down a step, drop a power”
The coefficient steps down (multiplied by the old power), and the power drops by one. If a term has no x at all, it drops to nothing — a constant differentiates to 0.
Products & quotients: expand first
A genuine trap: you cannot differentiate a product by differentiating each bracket and multiplying. Multiply the brackets out into a sum of powers, then differentiate.
Expand, then differentiate
(2x − 3)(x2 − 4) = 2x3 − 3x2 − 8x + 12
Never multiply the separate derivatives ✗
Same idea for quotients: split or simplify into separate power terms before applying the rule — e.g. 3x2 is just 3x−2, which differentiates to −6x−3.
Worked examples
The main SME example is f(x) = 2x3 + 4√x (with x > 0); the others reuse the SME illustrations.
WE 1
Rewrite f(x) = 2x3 + 4/√x as powers of x
Get every term into the form axn ready for the rule.
Apply the power rule to each term, taking care with the negative fractional power.
term by term: a n x^(n−1)2·3x² = 6x² ; 4·(−½)x^(−½−1) = −2x^(−3/2)careful: −½ − 1 = −3/2.f′(x) = 6x² − 2x^(−3/2)
WE 3
Differentiate y = −4x1/2
A single fractional-power term — a quick check of the rule.
y = −4x^(1/2)dy/dx = −4 · ½ · x^(½−1) = −2x^(−1/2)dy/dx = −2x^(−1/2)
WE 4
Differentiate f(x) = 5x4 − 3x2/3 + 4
A sum/difference — differentiate term by term, and don’t forget the constant.
5·4x³ − 3·(2/3)x^(2/3 − 1) + 0= 20x³ − 2x^(−1/3)f′(x) = 20x³ − 2x^(−1/3)the +4 is a constant, so it vanishes.
WE 5
Differentiate f(x) = (2x − 3)(x2 − 4)
A product — expand into powers of x first, then differentiate.
expand: 2x³ − 3x² − 8x + 12differentiate: 6x² − 6x − 8f′(x) = 6x² − 6x − 8multiplying the two separate derivatives would be wrong.
💡 Top tips
Rewrite before you differentiate — roots → fractional powers, fractions → negative powers. This single habit prevents most errors.
The rule is in the booklet (xn → nxn−1) — but the special cases ax → a and a → 0 are worth memorising.
Work term by term for any sum or difference; deal with each power independently.
Subtract 1 carefully from negative or fractional powers: −½ − 1 = −3/2, not −1/2.
Expand products/quotients into a sum of powers first — you cannot differentiate factor-by-factor.
Leave the form consistent with the question, or rewrite back into roots/fractions if asked.
⚠ Common mistakes
Differentiating a product factor-by-factor — multiplying the separate derivatives gives the wrong answer; expand first.
Forgetting to rewrite roots/fractions — leaving √x or 4x as-is means the rule can’t be applied.
Slips when subtracting 1 from negative or fractional powers.
Keeping the constant — a lone number differentiates to 0, not to itself.
Dropping the coefficient — remember to multiply the old power into the constant out front.
Next up — Gradients, Tangents & Normals. Now that you can write down f′(x) instantly, the next step is to use it: substitute an x-value to get the gradient at a point, then build the straight-line equations of the tangent (same gradient) and the normal (perpendicular, gradient −1/f′) through that point.
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