IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 ~9 min read

Differentiating Powers of x

In the last note we built up to the idea of a derivative using chords and limits. The good news? You’ll never have to do that the long way again. There’s a simple, mechanical rule for differentiating any power of x: bring the power down, then knock one off the top. Once it clicks, you’ll be finding derivatives in seconds.

📘 What you need to know

The power rule: if f(x) = xⁿ, then f′(x) = nx^{n − 1 (in formula booklet).}
If multiplied by a constant: f(x) = axⁿ → f′(x) = anx^{n − 1.}
Two special cases: (i) f(x) = ax → f′(x) = a · (ii) f(x) = a (constant) → f′(x) = 0.
The rule works for any rational power (fractions, negatives, etc.).
For roots, rewrite as fractional powers first: √x = x^1/2.
For fractions, rewrite as negative powers first: 1x = x⁻¹.
For sums/differences, differentiate term by term. Products and quotients must be expanded first.

The power rule

This is the single most important formula in the whole differentiation topic. Memorise it now and the rest of calculus becomes just bookkeeping.

The power rule

If f(x) = xⁿ then f′(x) = nx^{n − 1}

✓ in formula booklet

In words: bring the power down to the front, then reduce the power by 1. That’s the whole rule. Let’s see it in action with f(x) = x⁵:

x⁵power = 5 → 5x⁴5 came down · 5−1 = 4

“Bring it down, take one off”

🧠

Two-step chant: “Drop & drop”

Drop the power down to the front. Drop the power by 1. Two drops, every time. Once you say it out loud a few times, it becomes automatic.

What about a number in front?

If the power of x is multiplied by a constant, that constant just tags along through differentiation — it gets multiplied by the power as well.

Power rule with a constant

If f(x) = axⁿ then f′(x) = anx^{n − 1}

For example, if y = 3x⁴, then dydx = 3 × 4 × x³ = 12x³. The 3 stays put, while the 4 comes down and multiplies it.

Two special cases worth memorising

These come up so often that you don’t want to be working them out from the rule every time.

Linear term

f(x) = ax → f′(x) = a

e.g. y = 6x → dy/dx = 6

Constant term

f(x) = a → f′(x) = 0

e.g. y = 5 → dy/dx = 0

🤔 Why does a constant differentiate to zero?

A constant function is just a horizontal line — its height never changes as x moves. So its rate of change is zero. Anywhere on the graph, the gradient is flat (zero). The same logic explains why the constant in something like y = x² + 7 disappears when you differentiate — the +7 just shifts the curve up, but doesn’t affect how steep it is.

Rewrite first, differentiate second

The power rule only works on terms that look like axⁿ. So if a function has roots or fractions, the trick is to rewrite them as powers of x before applying the rule. Here’s the cheat sheet:

Original form	Rewrite as	Why
√x	x^1/2	square root = power of ½
³√x	x^1/3	cube root = power of ⅓
1x	x⁻¹	“1 over” = negative power
4x²	4x⁻²	denominator’s power becomes negative
1√x	x^−1/2	root in denominator → negative fraction

📍

The golden rule of differentiation

If it doesn’t look like axⁿ, rewrite it so it does. Every function in this topic can be written as a sum of terms in that form — once that’s done, the power rule handles the rest.

When dealing with negative or fractional powers, take extra care with the arithmetic. The power “−12 minus 1″ is “−32” — not “−12“. Slip-ups with negative signs are by far the most common mistake on this topic.

Sums and differences — differentiate term by term

If a function is built from several terms added or subtracted together, just differentiate each term separately and put it all back together. The order and signs stay the same.

Term-by-term rule

If f(x) = u(x) ± v(x) then f′(x) = u′(x) ± v′(x)

For example, with f(x) = 5x⁴ − 3x^2/3 + 4:

The 5x⁴ differentiates to 5 × 4 × x³ = 20x³.
The −3x^2/3 differentiates to −3 × 23 × x^−1/3 = −2x^−1/3.
The +4 (constant) differentiates to 0.

So f′(x) = 20x³ − 2x^−1/3.

Products and quotients — expand first!

This is where most students slip up. The power rule does not work directly on products like (2x − 3)(x² − 4) or quotients with x in the denominator (other than simple fractions). You need to multiply or simplify them out first, then differentiate term by term.

⚠️

You CAN’T differentiate a product by multiplying derivatives

The derivative of (x² + 3)(x³ − 2x + 1) is not the derivative of (x² + 3) times the derivative of (x³ − 2x + 1). Always expand the brackets out first, then differentiate term by term.

So for f(x) = (2x − 3)(x² − 4), first expand:

f(x) = 2x³ − 3x² − 8x + 12

Now it’s a sum/difference of powers. Differentiate term by term:

f′(x) = 6x² − 6x − 8

There is a “product rule” and a “quotient rule” for differentiating these directly — but at AA SL you only see them later (or sometimes not at all). At this stage, expanding first is the safe play. It also gets you method marks for showing your working.

Worked examples

WE 1

Apply the basic power rule

Differentiate the following with respect to x: (a) y = x⁷ (b) y = 4x³ (c) y = −2x⁵

Bring down the power, reduce by 1. The constant tags along.part (a) dy/dx = 7x⁶ 7x⁶part (b) dy/dx = 4 × 3 × x² = 12x² 12x²part (c) dy/dx = −2 × 5 × x⁴ = −10x⁴ −10x⁴ drop & drop! the negative sign just rides along.

WE 2

Sums, differences and special cases

Find dydx for y = 3x⁴ − 2x³ + 7x − 5.

Differentiate term by term. Watch the linear term and the constant. 3x⁴ → 12x³ −2x³ → −6x² 7x → 7 (linear special case) −5 → 0 (constant special case) dy/dx = 12x³ − 6x² + 7 the −5 vanishes — constants always disappear when differentiating!

WE 3

Roots and fractions — rewrite first

The function f(x) is given by f(x) = 2x³ + 4√x, where x > 0. Find f′(x).

The 4/√x doesn’t fit the power rule yet — rewrite it first as a power of x.step 1 — rewrite 4/√x = 4x^(−1/2) f(x) = 2x³ + 4x^(−1/2)step 2 — differentiate term by term 2x³ → 6x² 4x^(−1/2) → 4 × (−½) × x^(−1/2 − 1) = −2x^(−3/2) f′(x) = 6x² − 2x^(−3/2) careful with the new power: −½ − 1 = −³⁄₂ (not −½)!

WE 4

Products — expand first

Find dydx for y = (2x − 3)(x² − 4).

Can’t differentiate a product directly — expand the brackets first.step 1 — expand y = 2x · x² − 2x · 4 − 3 · x² + 3 · 4 y = 2x³ − 8x − 3x² + 12 y = 2x³ − 3x² − 8x + 12step 2 — differentiate term by term dy/dx = 6x² − 6x − 8 dy/dx = 6x² − 6x − 8 never multiply derivatives of brackets together — always expand first!

WE 5

A full mixed expression

Differentiate f(x) = 5√x − 3x² + 2x − 7 with respect to x.

Rewrite EVERY term as a power of x first, then apply the rule term by term.step 1 — rewrite 5√x = 5x^(1/2) 3/x² = 3x^(−2) f(x) = 5x^(1/2) − 3x^(−2) + 2x − 7step 2 — differentiate term by term 5x^(1/2) → 5 × ½ × x^(−1/2) = (5/2)x^(−1/2) −3x^(−2) → −3 × (−2) × x^(−3) = 6x^(−3) 2x → 2 −7 → 0 f′(x) = 52x^(−1/2) + 6x^(−3) + 2 two negatives in “−3 × −2” make a positive — sign tracking is everything!

💡 Top tips

“Bring down, take off one” — say it out loud the first few times until it’s automatic.
Always rewrite before you differentiate. Roots become fractional powers; fractions become negative powers.
Constants disappear, linear terms keep their coefficient. Memorise these special cases — they save real time.
Differentiate term by term for sums and differences. Don’t try to do it all in one line.
For products, expand the brackets first, then differentiate. Never multiply two derivatives together.
Watch the signs on negative powers. “n − 1” with n = −2 gives −3, not −1.
Show every step in working — the rewrite, then the differentiation. Examiners reward method marks for both.
You can leave answers in the form 6x² − 2x^(−3/2). You don’t have to convert back to roots and fractions unless told to.

⚠ Common mistakes

Trying to differentiate a product by multiplying derivatives. Always expand brackets first — this is the #1 error on this topic.
Forgetting to rewrite roots and fractions as powers of x. The power rule only works on terms in the form ax^n.
Sign errors with negative powers. If n = −2, then n − 1 = −3, not −1.
Forgetting that the constant disappears. Differentiating “+5” gives 0, not 5.
Forgetting that linear terms become constants. 7x differentiates to 7, not 7x.
Mishandling fractional powers. ½ − 1 = −½, not 0. Take care with the arithmetic.
Forgetting to multiply by both the original constant AND the power. 3x⁴ becomes 12x³ (3 × 4), not 4x³ or 3x³.
Differentiating each part of a quotient separately. If a fraction has x in the denominator, rewrite as a negative power first.

You’ve now got the power rule — the workhorse formula behind almost every Paper 1 calculus question. The next note shows how to use derivatives to find the gradient at a point, plus the tangent and normal lines to a curve. Same rule, more uses.

Need help with Differentiating Powers of x?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →

Differentiating Powers of x

📘 What you need to know

The power rule

Two-step chant: “Drop & drop”

What about a number in front?

Two special cases worth memorising

🤔 Why does a constant differentiate to zero?

Rewrite first, differentiate second

The golden rule of differentiation

Sums and differences — differentiate term by term

Products and quotients — expand first!

You CAN’T differentiate a product by multiplying derivatives

Worked examples

💡 Top tips

⚠ Common mistakes

Need help with Differentiating Powers of x?

Quick Links

Contact us

Follow us