IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
~7 min read
Differentiating Powers of x
The power rule turns differentiation into a one-line operation: bring the exponent down as a multiplier, then reduce the exponent by 1. With constants, sums and differences treated term by term, you can differentiate any polynomial — and any expression involving roots or reciprocal powers, after rewriting them as xn first.
📘 What you need to know
- Power rule: if f(x) = xn then f′(x) = nxn−1 for any rational n (booklet).
- With a constant coefficient: if f(x) = axn then f′(x) = anxn−1.
- Linear term: f(x) = ax → f′(x) = a (just the constant).
- Constant term: f(x) = a → f′(x) = 0 (constants vanish).
- Sums & differences: differentiate term by term — addition and subtraction commute with d/dx.
- Roots: rewrite √x as x1/2, ∛x as x1/3, etc., before differentiating.
- Fractions: rewrite 1/xk as x−k before differentiating.
- Products & quotients: expand or simplify FIRST — they don’t differentiate term by term.
The power rule
Power rule (with constant)
f(x) = axn ⇒ f′(x) = anxn−1
The mechanic: bring the exponent down as a multiplier, then subtract 1 from the exponent. The constant coefficient just rides along.
| f(x) | f′(x) | Comment |
|---|
| x5 | 5x4 | standard power |
| 3x2 | 6x | coefficient × exponent |
| 7x | 7 | linear → constant |
| −4 | 0 | constant → vanishes |
| x−2 | −2x−3 | sign flips on going down |
| x1/2 | 12x−1/2 | fractional exponent works the same |
Sums, differences, and constants
For an expression that’s a sum or difference of powers of x, just differentiate term by term:
Term-by-term differentiation
ddx(u(x) ± v(x)) = u′(x) ± v′(x)
This rule does NOT extend to products or quotients. For (3x − 2)(x² + 5x + 1) you can’t just differentiate each bracket — expand first, then differentiate term by term.
Roots and fractions — rewrite as xn first
The power rule needs the term in the form axn. Anything else has to be rewritten before differentiating.
Roots → fractional powers
√x = x1/2; ∛x = x1/3
general: k√(xm) = xm/k
Fractions → negative powers
1xk = x−k
remember the minus sign on the exponent
Sign-trap reminder: when reducing the exponent of a negative power, you SUBTRACT 1. So x−2 differentiates to −2x−3 (not −2x−1!) — the exponent goes from −2 to −3, more negative.
🧭 Recipe — differentiate any power-of-x expression
- Rewrite every term as a constant times a power of x (handle √, ∛, fractions).
- Apply the power rule to each term: bring the exponent down, subtract 1.
- Collect the resulting terms.
- Tidy up — convert back to root or fraction form if the question wants that style.
- Cross-check by computing f′ at one or two points on the GDC’s d/dx tool.
Worked examples
WE 1Polynomial — pure power rule, term by term
Differentiate f(x) = 5x⁴ + 3x² − 7x + 9.
Apply power rule to each term
d/dx[5x⁴] = 20x³
d/dx[3x²] = 6x
d/dx[−7x] = −7
d/dx[9] = 0
Collect
f′(x) = 20x³ + 6x − 7
f′(x) = 20x³ + 6x − 7
cross-check at x = 2: 20(8) + 12 − 7 = 165 ✓
WE 2Negative powers — rewrite reciprocals first
Differentiate g(x) = 2x³ + 5x.
Rewrite as negative powers of x
g(x) = 2x⁻³ + 5x⁻¹
Apply power rule
d/dx[2x⁻³] = 2(−3)x⁻⁴ = −6x⁻⁴
d/dx[5x⁻¹] = 5(−1)x⁻² = −5x⁻²
Collect (and convert back to fractions)
g′(x) = −6x⁻⁴ − 5x⁻² = −6/x⁴ − 5/x²
g′(x) = −6/x⁴ − 5/x²
careful with signs: −3 − 1 = −4 (more negative), not −2
WE 3Roots — rewrite as fractional powers first
Differentiate h(x) = 6√x + ∛x for x > 0.
Rewrite roots as fractional powers
h(x) = 6x^(1/2) + x^(1/3)
Apply power rule
d/dx[6x^(1/2)] = 6 · (1/2) · x^(−1/2) = 3x^(−1/2)
d/dx[x^(1/3)] = (1/3) · x^(−2/3)
Tidy up — convert to root form
h′(x) = 3/√x + 1/(3·∛(x²))
h′(x) = 3x^(−1/2) + (1/3)x^(−2/3)
subtracting 1 from a fraction: 1/3 − 1 = 1/3 − 3/3 = −2/3
WE 4Mixed positive, negative, fractional powers
Differentiate y = 4x³ − 2x² + 7√x for x > 0.
Rewrite every term as a power of x
y = 4x³ − 2x⁻² + 7x^(−1/2)
Apply power rule term by term
d/dx[4x³] = 12x²
d/dx[−2x⁻²] = −2(−2)x⁻³ = 4x⁻³
d/dx[7x^(−1/2)] = 7(−1/2)x^(−3/2) = −(7/2)x^(−3/2)
Collect
dy/dx = 12x² + 4x⁻³ − (7/2)x^(−3/2)
= 12x² + 4/x³ − 7/(2x√x)
dy/dx = 12x² + 4/x³ − 7/(2x√x)
at x = 1: 12 + 4 − 7/2 = 12.5 — quick sanity check passes
WE 5Product — expand FIRST, then differentiate
Differentiate f(x) = (3x − 2)(x² + 5x + 1).
Expand the product (you cannot differentiate each bracket separately)
3x · (x² + 5x + 1) = 3x³ + 15x² + 3x
−2 · (x² + 5x + 1) = −2x² − 10x − 2
Sum: f(x) = 3x³ + 13x² − 7x − 2
Apply power rule term by term
f′(x) = 9x² + 26x − 7
f′(x) = 9x² + 26x − 7
at x = 1: 9 + 26 − 7 = 28; differentiating each bracket separately would give 3 · (2x + 5) ≠ this
WE 6Multi-part: derivative function, value, and equation-solving
Let f(x) = x⁴ − 5x² + 8x − 3.
(a) Find f′(x). (b) Find f′(2). (c) Find the values of x for which f′(x) = 8.
(a) Differentiate term by term
f′(x) = 4x³ − 10x + 8
(b) Substitute x = 2
f′(2) = 4(8) − 10(2) + 8 = 32 − 20 + 8 = 20
(c) Set f′(x) = 8 and solve
4x³ − 10x + 8 = 8
4x³ − 10x = 0
2x(2x² − 5) = 0
→ x = 0 or 2x² = 5 → x² = 5/2 → x = ±√10/2
(a) f′(x) = 4x³ − 10x + 8; (b) f′(2) = 20; (c) x = 0, ±√10/2
factor out the common 2x — turns a cubic into a linear × quadratic
💡 Top tips
- Always rewrite first: roots become fractional powers, fractions become negative powers — THEN differentiate.
- Constants vanish on differentiation — don’t drag them along into your answer.
- For products and quotients, expand or simplify FIRST. You can’t differentiate (u)(v) by differentiating each factor separately.
- The exponent handle: bring it down, then reduce by 1. Both steps are non-negotiable.
- Cross-check on a GDC at one or two values of x using d/dx(…)|_x=…
⚠ Common mistakes
- Differentiating a product term by term — (x² + 3)(x³ − 2x + 1) doesn’t differentiate as (2x)(3x² − 2). Expand first.
- Forgetting to reduce the exponent — d/dx[x⁵] = 5x⁴, NOT 5x⁵.
- Sign error with negative exponents — x⁻² differentiates to −2x⁻³ (the exponent goes more negative).
- Treating √x directly — rewrite as x^(1/2) before differentiating; don’t memorise random “1/(2√x)” without seeing where it comes from.
- Dropping the coefficient — d/dx[3x²] = 6x, NOT 2x.
Next: Gradients, Tangents & Normals. Now that you can produce a derivative function, the immediate application is finding the gradient of a curve at any point — and using it to write the equation of the tangent (parallel to the curve at that point) or the normal (perpendicular to the tangent). It’s just “find a gradient, then plug into y − y₁ = m(x − x₁)”.
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