IB Maths AA SL Topic 5 โ€” Calculus Paper 1 & 2 ๐ŸŽฏ Skill ~4 min practice

AA SL Product Rule skills

When you have two functions multiplied together โ€” like xยฒ ยท sin x or ex ยท ln x โ€” you can’t differentiate them separately. The product rule glues them with a clean two-piece formula. Set up u and v in a small table and the rest is mechanical.

The Method

dydx  =  u dvdx  +  v dudx “first ร— derivative of second + second ร— derivative of first” ยท โœ“ in formula booklet
  1. Split the product โ€” call the first function u and the second v. Either order works.
  2. Differentiate both pieces separately โ€” find u‘ and v‘ in a small table.
  3. Plug into uv‘ + vu, then simplify (factor common terms if asked).

Set up u and v in a table

Example: differentiate y = xยฒ ยท sin x

u (first) v (second)
function xยฒ sin x
derivative 2x cos x
dy/dx = uv‘ + vu‘ = xยฒ cos x + 2x sin x

Build the table first, then read off the formula. No algebra, no surprises.

Product rule vs Chain rule

โœ“ USE product rule when

Two functions are multiplied: f(x) ยท g(x)

e.g. xยฒ sin x, ex ln x, (2x+1)(xโˆ’3)

Worked examples

WE 1 EASY

Differentiate y = xยณ(2x โˆ’ 5).

step 1 โ€” set up u and v u = xยณ   v = 2x โˆ’ 5 u’ = 3xยฒ   v’ = 2 step 2 โ€” apply uv’ + vu’ dy/dx = xยณ(2) + (2x โˆ’ 5)(3xยฒ) step 3 โ€” simplify = 2xยณ + 6xยณ โˆ’ 15xยฒ = 8xยณ โˆ’ 15xยฒ dy/dx = 8xยณ โˆ’ 15xยฒ could also expand first to 2xโด โˆ’ 5xยณ, then differentiate to get the same answer โ€” pick whichever is faster!
WE 2 MEDIUM

Differentiate y = 4x ยท cos x.

step 1 โ€” table u = 4x   v = cos x u’ = 4   v’ = โˆ’sin x step 2 โ€” apply formula dy/dx = 4x(โˆ’sin x) + (cos x)(4) dy/dx = 4 cos x โˆ’ 4x sin x d/dx(cos x) = โˆ’sin x โ€” keep the minus sign carefully when expanding!
WE 3 HARD

Differentiate y = xยฒ ยท e3x. Factor the answer.

step 1 โ€” set up u = xยฒ โ†’ u’ = 2x v = e3x โ†’ v’ = 3e3x (chain rule!) step 2 โ€” apply uv’ + vu’ dy/dx = xยฒ(3e3x) + e3x(2x) = 3xยฒe3x + 2xe3x step 3 โ€” factor common terms common factor: x ยท e3x dy/dx = x e3x(3x + 2) when “factor the answer” is asked โ€” pull out everything that’s common across both pieces!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Build the u/v table before plugging in โ€” saves errors every time.
Q1 EASY Differentiate y = x ยท ln x. Show answer โ–ผHide answer โ–ฒ
u = x โ†’ u’ = 1 v = ln x โ†’ v’ = 1/x dy/dx = x(1/x) + (ln x)(1) dy/dx = 1 + ln x
Q2 EASY Differentiate y = (2x + 1)(x โˆ’ 3). Show answer โ–ผHide answer โ–ฒ
u = 2x + 1 โ†’ u’ = 2 v = x โˆ’ 3 โ†’ v’ = 1 dy/dx = (2x + 1)(1) + (x โˆ’ 3)(2) = 2x + 1 + 2x โˆ’ 6 dy/dx = 4x โˆ’ 5 expanding first then differentiating is faster here โ€” both methods work!
Q3 MEDIUM Differentiate y = xยฒ ยท sin x. Show answer โ–ผHide answer โ–ฒ
u = xยฒ โ†’ u’ = 2x v = sin x โ†’ v’ = cos x dy/dx = xยฒ cos x + sin x ยท 2x dy/dx = xยฒ cos x + 2x sin x
Q4 MEDIUM Differentiate y = ex(xยฒ + 1). Show answer โ–ผHide answer โ–ฒ
u = ex โ†’ u’ = ex v = xยฒ + 1 โ†’ v’ = 2x dy/dx = ex(2x) + (xยฒ + 1)(ex) factor ex dy/dx = ex(xยฒ + 2x + 1) = ex(x + 1)ยฒ factoring reveals a perfect square hiding inside โ€” bonus simplification!
Q5 HARD Differentiate y = (3x โˆ’ 1)ยฒ ยท ln x. Factor the answer. Show answer โ–ผHide answer โ–ฒ
u and v setup (chain rule on u) u = (3x โˆ’ 1)ยฒ โ†’ u’ = 2(3x โˆ’ 1)(3) = 6(3x โˆ’ 1) v = ln x โ†’ v’ = 1/x apply uv’ + vu’ dy/dx = (3x โˆ’ 1)ยฒ(1/x) + ln x ยท 6(3x โˆ’ 1) factor (3x โˆ’ 1) dy/dx = (3x โˆ’ 1)[(3x โˆ’ 1)/x + 6 ln x] always look for the common factor โ€” it’s almost always there in IB questions!

โš  Common mistakes

๐Ÿ“–

Want the theory?

Read the full Product Rule notes for the proof, the link to the quotient rule, and worked exam-style problems combining product rule with chain rule.

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