IB Maths AI SLDifferentiationPaper 1 & 2f′(x) = anxn−1~6 min read
Differentiating Powers of x
Differentiation is the process of finding a function’s derivative — its gradient function — from a formula. For powers of x a single rule does the whole job. Learn it, add the two special cases, and you can differentiate any sum of powers term by term.
📘 What you need to know
The power rule: if f(x) = axn then f′(x) = anxn−1 — multiply by the power, then drop the power by 1.
A constant multiplier stays put: the derivative of axn is just a times the derivative of xn.
Special case 1: the derivative of ax is a — e.g. the derivative of 6x is 6.
Special case 2: the derivative of a constant is 0 — a constant has no gradient.
Differentiate a sum or difference of powers term by term.
Rewrite first: turn fractions like a/xn into ax−n, and expand products, before differentiating.
The power rule
Differentiation is the process of finding a formula for the derivative — the gradient function — of a function. For a power of x, one rule does it: multiply by the power, then reduce the power by 1. Any constant multiplier rides along unchanged.
The power rulef(x) = axn ⇒ f′(x) = anxn−1multiply by the power, then reduce the power by 1 · n is an integer
To differentiate axn: multiply the term by the power, then subtract 1 from the power. Below, −4x5 becomes −20x4.
This rule is given in the formula booklet, so you don’t need to memorise it — but you do need to apply it quickly and accurately.
Special cases and negative powers
Two short cases are worth knowing on sight. A term like 6x is really 6x1, and the rule turns it into 6x0 = 6. A constant on its own has no x to vary, so its gradient is 0.
Two special casesf(x) = ax ⇒ f′(x) = af(x) = a ⇒ f′(x) = 0a straight-line term ax has constant gradient a · a constant has gradient 0
The power rule also works for negative powers — but a fraction has to be rewritten first. A term like 4/x becomes 4x−1, and 5/x2 becomes 5x−2. Then differentiate as normal, taking care with the subtraction: reducing −2 by 1 gives −3.
Sums, products and quotients
An expression that is a sum or difference of powers is differentiated term by term — handle each piece separately, then add the results. For instance, the derivative of 5x4 + 2x3 − 3x + 4 is 20x3 + 6x2 − 3.
Products and quotients cannot be differentiated term by term as they stand. Expand a product of brackets, or simplify a quotient by dividing through, so that every term is a single power of x — only then does the power rule apply.
🧠Recipe — differentiating an expression
Rewrite every term as a power of x: turn a/xn into ax−n, and expand any brackets.
Take one term at a time — a sum or difference is differentiated piece by piece.
Apply the power rule: for each axn, multiply the coefficient by the power, then reduce the power by 1.
Use the special cases: a term ax becomes a; a constant becomes 0.
Write the derivative as the sum of the differentiated terms — convert negative powers back to fractions if asked.
Worked examples
WE 1
A single power-rule term
Find the derivative of y = 7x4.
multiply by the power 4, then reduce the power by 1dy/dx = 7 × 4 × x3dy/dx = 28x3the coefficient 7 just rides along — 7 × 4 = 28, and the power drops from 4 to 3.
WE 2
A polynomial, term by term
Differentiate f(x) = 3x5 − 6x2 + 9x − 11.
differentiate each term separately3x5 → 3×5 x4 = 15x4−6x2 → −6×2 x = −12x9x → 9 · −11 → 0f′(x) = 15x4 − 12x + 9the +9 comes from the 9x term, and the −11 vanishes — every constant differentiates to 0.
WE 3
A negative power from a fraction
Find f′(x) for f(x) = 2x3 + 5/x2.
rewrite the fraction as a negative power: 5/x2 = 5x−2f(x) = 2x3 + 5x−22x3 → 6x25x−2 → 5×(−2) x−3 = −10x−3f′(x) = 6x2 − 10x−3 = 6x2 − 10/x35/x2 must become 5x−2 before differentiating. Reducing the power −2 by 1 gives −3.
WE 4
Mixed: negative power and constant
Find dy/dx for y = 4x − 3/x + 6.
rewrite: 4x − 3x−1 + 64x → 4 · 6 → 0−3x−1 → −3×(−1) x−2 = 3x−2dy/dx = 4 + 3x−2 = 4 + 3/x2−3 × (−1) = +3, so the sign flips — a common slip when differentiating negative powers.
WE 5
A product — expand first
Find f′(x) for f(x) = (x + 4)(2x − 1).
a product — expand before differentiating(x + 4)(2x − 1) = 2x2 − x + 8x − 4f(x) = 2x2 + 7x − 4f′(x) = 4x + 7you cannot multiply the derivatives of the two brackets — expand to a sum of powers first.
WE 6
Full question: quotient, then a gradient
A curve has equation y = (4x3 + 2x)/(2x). (a) Show that y = 2x2 + 1. (b) Find dy/dx. (c) Find the gradient of the curve at x = 3.
(a) divide each term of the numerator by 2x4x3 ÷ 2x = 2x2 · 2x ÷ 2x = 1⇒ y = 2x2 + 1(b) differentiate: 2x2 → 4x, 1 → 0dy/dx = 4x(c) gradient at x = 3: substitute into dy/dxdy/dx = 4 × 3(a) y = 2x2+1 · (b) dy/dx = 4x · (c) gradient = 12a quotient must be simplified to a sum of powers first. The gradient at a point comes from substituting that x-value into the derivative.
💡 Top tips
The power rule has two moves in order: multiply by the power, then reduce the power by 1 — don’t skip the second.
Rewrite fractions as negative powers before differentiating: a/xn becomes ax−n.
The derivative of a constant is 0, and the derivative of ax is a — quick marks once you spot them.
Expand products and simplify quotients first — they can’t be differentiated term by term as they stand.
Take care with negative powers: reducing −2 by 1 gives −3, not −1.
âš Common mistakes
Forgetting to reduce the power — writing the derivative of x4 as 4x4 instead of 4x3.
Differentiating a product by multiplying the separate derivatives — expand the brackets first.
Leaving a constant in the derivative — the derivative of a plain number is 0.
Mishandling negative powers: the derivative of x−1 is −x−2, since −1 − 1 = −2.
Dropping the coefficient — the derivative of 5x3 is 15x2, not 3x2.
Next up: Gradients, Tangents & Normals — putting the power rule to work to find the gradient at a point and the equations of tangent and normal lines. The golden rule before you differentiate anything: get every term into the form axn first. Expand brackets, split fractions — then the power rule does the rest.
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