IB Maths AA SL
Topic 5 — Calculus
Paper 1 & 2
~7 min read
Quotient Rule
Use the quotient rule when one function is divided by another — like sin xx or x2ex. Looks intimidating, but it’s just one formula. The order matters because of the minus sign.
📘 What you need to know
- The rule: if y = uv, then dydx = vu′ − uv′v2 (in formula booklet).
- Use it when both numerator AND denominator are functions of x.
- Order matters: vu′ comes first, then minus uv′.
- Don’t forget the v2 on the bottom.
- If only the numerator is constant, rewrite as a negative power and use chain rule instead — quicker.
Product or quotient? Spot the difference
Product rule
x · sin x
two functions multiplied
Quotient rule
sin x / x
one function divided by another
The formula
The 2-step method
How to apply the quotient rule
- Identify u (top) and v (bottom), then differentiate to get u′ and v′.
- Substitute into y′ = (vu′ − uv′)/v². Be careful with the order.
The square trick
Same 2×2 layout as product rule — but now with a minus instead of a plus, and divided by v2.
The quotient rule layout
y′ = (v · u′ − u · v′) / v2
v · u′ comes FIRST — order matters!
🧠“Lo d-Hi minus Hi d-Lo, all over Lo squared”
Lo = bottom, Hi = top, d = “derivative of”. So: bottom × derivative-of-top, minus top × derivative-of-bottom, all divided by the bottom squared. Sing it once and it sticks.
📍If the top is just a constant, skip the quotient rule
For something like 2(3x − 7)2, rewrite as 2(3x − 7)−2 and use the chain rule. Faster, fewer mistakes.
Worked examples
WE 1Polynomial / polynomial
Differentiate y = (x² + 1) / (x − 3).
step 1 — set up
u = x² + 1, v = x − 3
u′ = 2x, v′ = 1step 2 — apply formula
y′ = ((x − 3)(2x) − (x² + 1)(1)) / (x − 3)²
= (2x² − 6x − x² − 1) / (x − 3)²
y′ = (x² − 6x − 1) / (x − 3)²
expand the top first, then collect like terms — keep the bottom factored!
Differentiate y = sin x / x.
step 1
u = sin x, v = x
u′ = cos x, v′ = 1step 2
y′ = (x cos x − sin x · 1) / x²
y′ = (x cos x − sin x) / x²
classic AA SL question — no further simplification needed!
WE 3With chain rule inside
Differentiate f(x) = cos 2x / (3x + 2).
step 1 — chain rule for u′
u = cos 2x, v = 3x + 2
u′ = −2 sin 2x (chain rule)
v′ = 3step 2
f′(x) = ((3x+2)(−2 sin 2x) − (cos 2x)(3)) / (3x+2)²
f′(x) = (−2(3x+2) sin 2x − 3 cos 2x) / (3x+2)²
factor out −1 if needed: −((2(3x+2) sin 2x + 3 cos 2x)) / (3x+2)²
WE 4Polynomial / exponential
Differentiate y = x² / ex.
step 1
u = x², v = ex
u′ = 2x, v′ = exstep 2
y′ = (ex(2x) − x²(ex)) / (ex)²
= ex(2x − x²) / e2x
= (2x − x²) / ex
y′ = x(2 − x) / ex
ex cancels: ex/e2x = 1/ex. Always look for these!
WE 5Gradient at a specific point
Find the gradient of y = x / (x² + 1) at x = 2.
step 1
u = x, v = x² + 1
u′ = 1, v′ = 2xstep 2 — derivative
y′ = ((x² + 1)(1) − x(2x)) / (x² + 1)²
= (x² + 1 − 2x²) / (x² + 1)²
= (1 − x²) / (x² + 1)²step 3 — sub x = 2
y′(2) = (1 − 4) / (4 + 1)² = −3/25
gradient = −3/25
simplify the algebra BEFORE plugging in numbers!
💡 Top tips
- v u′ comes FIRST — bottom × derivative-of-top. Then minus u v′.
- Always square the bottom. Don’t forget the v² in the denominator.
- Use chain rule when needed for u′ or v′ (e.g. cos(2x) needs chain rule).
- If the top is a constant, rewrite as a negative power and use chain rule — quicker.
- Simplify first, then substitute for “find the gradient at x = a” questions.
- Factor the numerator if the question asks for a tidy form.
⚠ Common mistakes
- Wrong order: writing uv′ − vu′ instead of vu′ − uv′. The minus sign means order matters.
- Forgetting v² on the bottom.
- Sign errors when expanding −u(v′). A minus sign in front of a bracket flips every term inside.
- Skipping chain rule when finding u′ or v′ for composite expressions.
- Using product rule by accident for divisions. Sketch the formula in your head first.
You’ve now got chain, product, and quotient — the three big rules. Next we look at second order derivatives — what you get by differentiating twice, and why that’s so useful.
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